Time t $$\infty $$ Rotation of Glucose & Fructose $$ r_{t} $$ $$r _{\infty }$$ $$S\rightarrow$$ $$G+F $$ What id the value of $$k$$ for the above reaction under given circumstances?

$$\displaystyle k=\frac{1}{t}ln\frac{r_{t}}{(r_{\infty }-r_{t})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{r_{\infty }}{(r_{t}-r_{\infty})}$$

$$\displaystyle k=\frac{1}{t}ln\frac{r_{t}}{(r_{t}-r_{\infty})}$$

MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 10

In the reaction,

$2N_2O_5\rightarrow 4NO_2+O_2$ , initial pressure is

$500\ atm$ and rate constant

$k$ is

$3.38\times 10^{-5} sec^{-1}$ . After 10 minutes the final pressure of

$N_2O_5$ is:

0%
490 atm
0%
250 atm
0%
480 atm
0%
420 atm

Explanation

$2N_2O_5\rightarrow 4NO_2+O_2$ ,

$p_0$

$p_0-2x$

$4x$

$x$

$k=\dfrac {2.303}{t} log_{10}\dfrac {p_0}{p_1}$

$3.38\times 10^{-5}=\dfrac {2.303}{10\times 60} log \dfrac {500}{p_t}$

$p_t=490\ atm$

Hence, option A is correct.

$A(g)\rightarrow B(g)+C(g)$

$\dfrac {-d[A]}{dt}=k[A]$

At the start, the pressure is 100 mm and after 10 min, the pressure is 120 mm. Hence, rate constant

$(min^{-1})$ is:

0%
$\dfrac {2.303}{10} log (\dfrac {120}{100})$
0%
$\dfrac {2.303}{10} log (\dfrac {100}{20})$
0%
$\dfrac {2.303}{10} log (\dfrac {100}{80})$
0%
$\dfrac {2.303}{10} log( \dfrac {100}{120})$

Explanation

$A(g)\rightarrow B(g) + C(g)$

$t=0$ P 0 0
At time t P-x x x

Total pressure

$=P-x+x+x=120$ .

$P+x=120$

$x=120-P=120-100=20.$

$k=\dfrac {2.303}{t}log \dfrac {a}{a-x}$

$\ k=\dfrac {2.303}{10} log \dfrac {100}{80}$

Hence, option C is correct.

For the

$1^{st}$ order reaction:

$A_{(g)}\rightarrow 2B_{(g)}+C_{(s)}$ , the value of

$t_{\frac {1}{2}}=24\ mins$ . The reaction is carried out by taking a certain mass of

$A$ enclosed in a vessel in which it exerts a pressure of

$400\ mm\ Hg$ . The pressure of the reaction mixture after the expiry of

$48\ mins$ will be:

0%
$700\ mm$
0%
$600\ mm$
0%
$800\ mm$
0%
$1000\ mm$

In the first order reaction the concentration of reactant decreases from

$1.0\ M$ to

$0.25\ M$ in

$20$ minutes. The value of specific rate is:

0%
$69.32$
0%
$6.932$
0%
$0.6932$
0%
$0.06932$

Explanation

$\frac {2.303}{20} log \frac {2}{0.5}=0.06932$ .

Calculate the half-life of the first-order reaction,

$C_2H_4O(g)\rightarrow CH_4(g)+CO(g)$ , if the initial pressure of

$C_2H_4O(g)$ is 80 mm and the total pressure at the end of 20 minutes is 120 mm.

0%
40 min
0%
120 min
0%
20 min
0%
80 min

Explanation

For first-order reaction,

$k=\dfrac {2.303}{t} log \dfrac {a}{a-x}$

$=\dfrac {2.303}{20}\times log\dfrac {80}{40}$

$=\dfrac {2.303}{20}\times 0.3010$

For half-life period,

$t_{\frac {1}{2}}=\dfrac {0.693}{k}=\dfrac {0.693\times 20}{2.303\times 3010}=20\ min$

Hence, option C is correct.

A reaction that is of the first order with respect to reactant A has a rate constant 6 min

$^{-1}$ . If we start with [A]=0.5 mol L

$^{-1}$ , when would [A] reach the value of 0.05 mol L

$^{-1}$ ?

0%
0.384 min
0%
0.15 min
0%
3 min
0%
3.84 min

Explanation

We know that for first order kinetics:

$k=\dfrac {2.303}{t} log \dfrac {a}{a-x}$

$t=\dfrac {2.303}{6} log \dfrac {0.5}{0.05}=\dfrac {2.303}{6}=0.384\ min$

Hence, option A is correct.

For the first order reaction

$A_{(g)}\rightarrow 2B_{(g)}+C_{(g)}$ , the initial pressure is

$P_A=90 mm$ Hg, the pressure after 10 minutes is found to be 180 mm Hg. The rate constant of the reaction is:

0%
$1.15\times 10^{-3} sec^{-1}$
0%
$2.3\times 10^{-3} sec^{-1}$
0%
$3.45\times 10^{-3} sec^{-1}$
0%
$6\times 10^{-3} sec^{-1}$

Explanation

$A\rightarrow 2B+C$
P 0 0
P-x 2x x
At equilibrium,

$180=P-x+2x+x$

$180=90+2x$

$2x=90, x=45$

$k=\dfrac {2.303}{t} log \dfrac {P}{P-x}$

$=\dfrac {2.303}{10} log\dfrac {90}{90-45}=\dfrac {2.303}{10} log 2=\dfrac {0.6932}{10}$

$=0.6932=\dfrac {0.06932}{60}=1.1555\times 10^{-3} sec^{-1}$ .

The rate constant

$(k)$ of a first-order reaction is

$0.0693\ min^{-1}$ . If we start with

$20\ mol\ L^{-1}$ , then it is reduced to

$2.5\ mol\ L^{-1}$ in:

0%
$10\ mins$
0%
$20\ mins$
0%
$30\ mins$
0%
$40\ mins$

A certain zero order reaction has

$k=0.025\ M sec^{-1}$ for the disappearance of A. What will be the concentration of A after 15 seconds if initial concentration is 0.5 M?

0%
$0.5 \ M$
0%
$0.32 \ M$
0%
$0.125 \ M$
0%
$0.06 \ M$

Explanation

$x=kt=0.025\times 15=0.375\ M$
Remaining concentration of

$A= 0.5-0.375$

$=0.125\ M$

In the first order reaction, the concentration of the reactant is reduced to 25% in one hour.The half-life period of the reaction is:

0%
2 hr
0%
4 hr
0%
1/2 hr
0%
1/4 hr

Explanation

$k=\frac {2.303}{t} log \frac {a}{a-x}$

$k=\frac {2.303}{1} log \frac {100}{25}=\frac {2.303}{1} log 4=2.303\times 2\times 0.3010$

$t_{\frac {1}{2}}=\frac {0.693}{k}\therefore t_{\frac {1}{2}}=0.5 hr$ .

In the following gaseous phase first order reaction

$A(g) \rightarrow 2 B(g) + C(g)$
initial pressure was found to be

$400$ mm of

$Hg$ and it changed to

$1000$ mm of

$Hg$ after $$20$4 min. Then:

0%
half life for $A$ is $10$ min
0%
rate constant is $0.0693$ min $^{-1}$
0%
partial pressure of $C$ at $30$ min is $350$ mm of $Hg$
0%
total pressure after $30$ min is $1100$ mm of $Hg$

Explanation

A(g)

$\rightarrow$ 2B(g) + C(g)
t = 0 400 0 0
t = 20 min 400 - p 2p p
Given

$400 - p + 2p + p = 1000$

$400 + 2p = 1000$

$p = 300 mm$

$k = \frac{1}{20}ln \frac{400}{400-300} = \frac{1}{20} ln 4$

$k = frac{ln 2}{10} min^{-1}$

$T_{1/2} = 10 min$

$Value of K = 0.0693 min^{-1}$
After 30 min Partial Pressure of A is 50 mm
After 30 min Partial Pressure of B is 700 mm
After 30 min Partial Pressure of C is 350 mm
After 30 min total pressure become 1100 mm
So all of the above are correct.

For a 1

$^{st}$ order reaction (gaseous) (constant V, T)

$a A \rightarrow (b - 1) B + 1 C$ (with b > a) the pressure of the system rose by

$50 \left ( \frac{b}{a} - 1 \right )$ % in a time of 10 min. The half life of the reaction is therefore:

0%
10 min
0%
20 min
0%
30 min
0%
40 min

Explanation

$aA \rightarrow$ (b -1) B + C

$t = 0;$

$P_0$ 0 0

$t= t;$

$P_0 - x$

$\frac{(b-1)x}{a}$

$\frac{x}{a}$

Total pressure

$=P_0 - x + \frac{b}{a} x = P_0 + \left ( \frac{b}{a} - 1 \right ) \frac{1}{2} P_0$

$P_0 + \left ( \frac{b}{a} - 1 \right ) x = P_0 + \left ( \frac{b}{a} - 1 \right ) \frac{P_0}{2}$

$x = \frac{P_0}{2}$

$P_A = P_0 - x = P_0 - \frac{P_0}{2} = \frac{P_0}{2}$

$P_A$ reduces to half in 10 min.

So,

$t_{1/2} = 10\ min.$

The half-period T for the decomposition of ammonia on tungsten wire was measured for different initial pressures P of ammonia at 25

$^o$ C. Then:

P(mm Hg)	11	21	48	73	120
T(sec)	48	92	210	320	525

0%
Zero order reaction
0%
First order reaction
0%
Rate constant for reaction is 0.114 mol lit $^{-1}$ sec $^{-1}$
0%
Rate constant for reaction is 1.14 seconds

Explanation

For nth ordr reaction,

$\displaystyle t_{1/2} \alpha (a)^{1-n} or (1-n) = \frac{log t'_{1/2} - log t"_{1/2}}{log a' - log a"}$
In terms of partial pressures,

$t_{1/2} \alpha \frac{1}{p^{n-1}}$
Substitute values in the above expression.

$\frac{21}{11} = \left ( \frac{49}{92} \right )^{n-1} n = 0$
Thus the reaction is of zero order.
The expression for the rate constant is

$t_{1/2} = \frac{P}{2K}$ or

$K = \frac{P}{2 t_{1/2}}$

$K = \frac{11}{2 \times 48} = 0.114 sec.$ .
Thus, the rate constant of the reaction is

$0.114 sec$

The energy of activation for the backward reaction is :

0%
$30 kJ mol^{-1}$
0%
$20 kJ mol^{-1}$
0%
$10 kJ mol^{-1}$
0%
$40 kJ mol^{-1}$

Explanation

Explanation:

Activation energy is the minimum energy that the reactants should possess to undergo a reaction.
Relation between heat of reaction and the energy of activation of forward and back reaction is;

$\Delta H_{reaction} = E_{a \ backward} - E_{a \ forward}$

From the above graph, we can find out the Enthalpy of reaction and Activation energy for forward reaction.

Since,

$\Delta H_{of reaction}$ =

$10 \ kJmol^{-1}$ and

$E_{Forward Reaction}$ =

$30 \ kJmol^{-1}$

$\Delta H_{reaction} = E_{a \ backward} - E_{a \ forward}$
Putting the values in the equation we will get;

$10 \ kJmol^{-1} =$ $E_{a \ backward} - 30 \ kJmol^{-1}$
$E_{a \ backward}$ $= 30 \ kJmol^{-1} + 10 \ kJmol^{-1}$
$E_{a \ backward}$ $= 40 \ kJmol^{-1}$

Therefore, $E_{Activation}$ for backward reaction = $40 \ kJmol^{-1}$

Hence, option D is the correct answer.

Which of the following statements about zero order reaction is/are not true?

0%
Unit of rate constant is $sec^{-1}$ .
0%
The graph between log (reactant) versus time is a straight line.
0%
The rate of reaction increases with the increase in concentration of reactants.
0%
Rate of reaction is independent of concentration of reactants.

Explanation

Let us consider a zero order reaction:

$A \ \rightarrow B$

Rate Law for Zero Order Reaction is given as :

$Rate \ = \ -\cfrac{d[A]}{dt} \ = \ k[A]^0 \ = \ k$

where k is Rate Constant

Hence, The unit of rate constant is the unit of Rate of Reaction i.e.

$mol \ litre^{-1} \ sec^{-1}$

$\because \ -\cfrac{d[A]}{dt} \ = \ k$

$\implies \ [A] = [A_0] - kt$ . . . . (1)

where

$[A_0]$ is the initial concentration of Reactant A.

taking log in the equation (1), we get,

$log\, [A] \ =\ log\,[A_0] - log[kt]$ . . . . (2)

$\because \ (2) \ is \ not \ of \ form \ log\, [A] = mt \ . \ \therefore \text{The graph between log[Reactant] versus time is not a straight line.}$

$\because \text{Rate of Zero order reaction is equal to Rate constant of that reaction, which is independent of concentration of reactant.}$

Hence, on increasing concentration of reactants, rate of reaction does not increase.

From Rate Law for Zero Order reaction, we can conclude that rate of reaction is independent of concentration of reactants.

Hence, Option

$"A", \ "B" \ \& \ "C"$ are not true.

Which of the following reactions is/are of the first order?

0%
The decomposition of ammonium nitrate in an aqueous solution.
0%
The inversion of cane-sugar in the presence of an acid.
0%
The acidic hydrolysis of ethyl acetate.
0%
All radioactive decays.

Explanation

The rate of first order reaction depends only on one concentration term of reactant. A first order reaction is one whose rate varies as first power of the concentration of the reactant, i.e. the rate increases as number of times as the concentration of reactant is increased.

Examples are as follows:

Decomposition of ammonium nitrate in aqueous solution.

$NH_4NO_2 \rightarrow N_2+2H_2O$

Inversion of cane sugar.

$C_{12}H_{22}O_{11}+H_2O \xrightarrow {[H^+]} C_6H_{12}O_6 +C_6H_{12}O_6$

Hydrolysis of ethyl acetate and radioactive deacys are other examples of first order reactions.

For the first order reaction,

$A(g)\rightarrow 2B(g)+C(g)$ , the initial pressure is

$P_A=90$ mm Hg. Then pressure after 10 minutes is found to be 180 mm Hg. The half-life period of the reaction is:

0%
$1.15\times 10^{-3} sec^{-1}$
0%
$600\;sec$
0%
$3.45\times 10^{-3} sec^{-1}$
0%
$200\;sec$

Explanation

$A\rightarrow 2B+C$

$P$

$0$

$P-x$

$2x$

$x$
Total pressure after 10 minute is

$180=P-x+2x+x$

$180=90+2x$

$2x=90, x=45$

$k=\dfrac {2.303}{t}log\dfrac {P}{P-x}$

$=\dfrac {2.303}{10} log\dfrac {90}{90-45}$

$=\dfrac {2.303}{10\times 60} log2$

$=1.1555\times 10^{-3} sec^{-1}$

$\therefore t_{\frac {1}{2}}=0.692/k$

$\approx 600 sec$

The initial rate of zero-order reaction of the gaseous equation

$A$ (g)

$\rightarrow$

$2B$ (g) is 10

$^{-2}$ M min

$^{-1}$ if the initial concentration of

$A$ is 0.1 M. What would be a concentration of

$B$ after

$60$ seconds?

0%
$0.09 M$
0%
$0.01 M$
0%
$0.02 M$
0%
$0.03 M$

Explanation

For a zero order reaction, the rate law expression is

$R = k[A]^0=k=10^{-2}$

$t=60 s = 1 min$

$[A_{0}] - [A] = Kt$

$0.1 - [A] = 10^{-2} \times 1$

$[A]=0.09 M$

$[B]=2[A]$

$=2(0.1-0.09)$

$= 0.02 M$

In 20 minutes of 80% of

$N_2O_5$ is decomposed. Rate constant is:

0%
0.08
0%
0.05
0%
0.12
0%
0.2

Explanation

$k=\frac {2.303}{t} log \frac {A_0}{A_t}=\frac {2.303}{20} log \frac {100}{20}$
On solving

$k=0.08$

The inactivation of a viral preparation in a chemical bath is found to be a first-order reaction. The rate constant for the viral inactivation per minute, if in the beginning

$1.5\%$ of the virus is inactivated, is:

0%
$1.25\times 10^{-4}\ sec^{-1}$
0%
$2.5\times 10^{-4}\ sec^{-1}$
0%
$5\times 10^{-4}\ sec^{-1}$
0%
$2.5\times 10^{-4}\ min^{-1}$

A certain substance A is mixed with an equimolar quantity of substance B. At the end of an hour, A is 75% reacted. Calculate the time when A is 10% unreacted. (Given: order of reaction is zero)

0%
1.2 hr
0%
1.8 hr
0%
2.0 hr
0%
2.4 hr

Explanation

According to zero order reaction:

$x =kt$ ( x = amount of reactant reacted)

$75 = k\times1$ and

$90 = k\times t$
So

$t= 90/75 =1.2$ hr

A drug was known to be effective after it has decomposed 30 %. The original concentration of a sample was 500 units/ml. When analyzed 20 months later, the concentration was found to be 420 units/ml. Assuming that decomposition is of Ist order, what will be the expiry time of the drug?

0%
41 months
0%
40 months
0%
35 months
0%
38 months

Explanation

For a first order reaction expression for the specific rate constant is

$\mathrm{k}=\dfrac{2.303}{\mathrm{t}}\log\dfrac{\mathrm{a}}{(\mathrm{a}-\mathrm{x})}$ .
20 months later, the concentration was found to be 420 units/ml.

$\mathrm{k}=\dfrac{2.303}{\mathrm{20}}\log\dfrac{\mathrm{500}}{(\mathrm{420})}$ ......(1)

For 30 % decomposition,

$\mathrm{k}=\dfrac{2.303}{\mathrm{t}}\log\dfrac{\mathrm{500}}{(\mathrm{500}-\mathrm{150})}$ ......(2)
From (1) and (2),

$\dfrac{2.303}{\mathrm{20}}\log\dfrac{\mathrm{500}}{(\mathrm{420})}=\dfrac{2.303}{\mathrm{t}}\log\dfrac{\mathrm{500}}{(\mathrm{500}-\mathrm{150})}$

On calculation, we get-

$t=41 \ months$

The rate constant for disappearance of reactant as mentioned in is

$\displaystyle 2 \times 10^{-2} mol L^{-1} sec^{-1}$ , if the concentration of the reactant after 25 sec is 0.25M, the initial concentration is:

0%
0.75 M
0%
0.89 M
0%
1.03 M
0%
1.46 M

Explanation

From rate constant it can be seen that it is zero order reaction and

$a-x = kt;$

$a = x+kt$

$= 0.25+2 \times 10^{-2} \times 25 = 0.75$ M.

A first order reaction is 20% complete in 10min. The specific rate constant is:

0%
0.0223 $min^{-1}$
0%
0.0423 $min^{-1}$
0%
0.0501 $min^{-1}$
0%
0.0517 $min^{-1}$

Explanation

A first order reaction is 20% complete in 10min.
The expression for the specific rate constant is

$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log\frac{\mathrm{a}}{(\mathrm{a}-\mathrm{x})}=\frac{2.303}{\mathrm{10}}\log\frac{\mathrm{100}}{(\mathrm{100}-\mathrm{20})}=0.0223 min^{-1}$ .

For the first order homogeneous gaseous reaction

$A\rightarrow 2B+C$ , the initial pressure was

$P_i$ while total pressure at time 't' was

$P_t$ . Write expression for the rate constant k in terms of

$P_i, P_t$ &

$t$ .

0%
$k=\cfrac {2.303}{t}log \left (\cfrac {2P_i}{3P_i-P_t}\right )$
0%
$k=\cfrac {2.303}{t}log \left (\cfrac {2P_i}{ 2P_t-P_i}\right )$
0%
$k=\cfrac {2.303}{t}log \left (\cfrac {P_i}{P_i-P_t}\right )$
0%
None of these

Explanation

The first-order reaction is

$A\rightarrow 2B+C$

	A	B	C
At time 0:	Pi	0	0
At time t:	Pi−x	2x	x

Hence, the expression for the rate constant will be:

$K=\dfrac {2.303}{t}log \left(\cfrac {P_i}{P_i-x}\right)$

But

$P_i-x+2x+x=P_t$

Hence,

$x=\cfrac {1}{2}(P_t-P_i)$ or

$P_i-x=\cfrac {1} {2}(3P_i-P_t)$

Thus the expression for the rate constant becomes

$K=\dfrac {2.303}{t} log \left(\cfrac {2P_i}{(3P_i-P_t)}\right)$

Hence, the correct option is

$\text{A}$

A viral preparation was inactivated in a chemical bath. The inactivation process was found to be first order in virus concentrations. At the beginning of the experiment, 2.0% of the virus was found to be inactivated per minute. The k for inactivation process is:

0%
$3\times10^{-4} s^{-1}$
0%
$4\times10^{-4} s^{-1}$
0%
$5\times10^{-4} s^{-1}$
0%
$6\times10^{-4} s^{-1}$

Explanation

2.0% of the virus was found to be inactivated per minute. The expression for the specific rate constant is

$\mathrm{k}=\frac{2.303}{\mathrm{t}}\log\dfrac{\mathrm{a}}{(\mathrm{a}-\mathrm{x})}=\dfrac{2.303}{\mathrm{1 \times 60}}\log\dfrac{\mathrm{100}}{(\mathrm{100}-\mathrm{2})}=3\times10^{-4} s^{-1}$ .

A solution of

$A$ is mixed with an equal volume of a solution of

$B$ containing the same number of moles, and the reaction

$A+B \longrightarrow C$ occurs. At the end of 1 hour,

$A$ is

$75$ % reacted. The amount of

$A$ that will be left unreacted at the end of

$2$ hours if the reaction is first order in

$A$ and zero order in

$B$ is:

0%
$6.25$
0%
$7.28$
0%
$8.43$
0%
$8.92$

Explanation

The integrated rate law for the first order reaction is

$k=\dfrac {2.303} {t} log \dfrac{[A]_0} {[A]_t}$ .

For

$t=1 \ hr$ ,

$k=\dfrac {2.303} {1} log \dfrac{{100}}{ {(100-75)}}$ -------- (1)

For

$t=2 \ hr$ ,

$k=\dfrac {2.303} {2} log \dfrac{ {100} }{{[A]_t}}$ -------- (2)

From equations (1) and (2),

$\dfrac {2.303} {1} log \dfrac{{100}}{ {(100-75)}}=\dfrac {2.303} {2} log \dfrac{{100} }{{[A]_t}}$

Hence,

$[A]_t=6.25$

For the following data for the zero order reaction

$A\rightarrow$ products. Calculate the value of k.
Time [A]
0.0 0.10 M
1.0 0.09 M
2.0 0.08 M

0%
$k=0.01 \, M min^{-1}$
0%
$k=0.03 \, M min^{-1}$
0%
$k=0.06 \, M min^{-1}$
0%
$k=0.08 \, M min^{-1}$

Explanation

For zero order reaction:

$rate = k$

$k = (0.10-0.09)/1 = 0.01/1 = 0.01 M min^{-1}$

$Rate = \dfrac{0.09-0.08}{2-1}= 0.01 M min^{-1}$

The decomposition of

${\mathrm{N}_{2}}\mathrm{O}_{5}$ according to the equation

${\mathrm{2N}_{2}}\mathrm{O}_{5}(g)\rightarrow \mathrm{4 NO}_{2}+O_{2}(g)$ is a first-order reaction. After 30 min from the start of decomposition in a closed vessel the total pressure developed is found to be 284.5 mm Hg. On complete decomposition, the total pressure is 584.5 mm Hg. The rate constant of the reaction is :

0%
$\mathrm{k}_{1} =5.206\times10^{-3}\ \mathrm{min}^{-1}$
0%
$\mathrm{k}_{1} =3.102\times10^{-3}\ \mathrm{min}^{-1}$
0%
$\mathrm{k}_{1} =3.126\times10^{-3}\ \mathrm{min}^{-1}$
0%
$\mathrm{k}_{1} =3.453\times10^{-3}\ \mathrm{min}^{-1}$

Explanation

$2{ \mathrm{N} }_{ 2 }{ \mathrm{O} }_{ 5 }\longrightarrow \mathrm{4N}{ \mathrm{O} }_{ 2 }+{ \mathrm{O} }_{ 2 }$

On decomposition of

$2$ moles of

${ \mathrm{N} }_{ 2 }{ \mathrm{O} }_{ 5 }$ ,

$4$ moles of

$\mathrm{N}{ \mathrm{O} }_{ 2 }$ and

$1$ mole of

${ \mathrm{O} }_{ 2 }$ are produced. Thus, the total pressure after completion corresponds to

$5$ moles and initial pressure to

$2$ moles.

Initial pressure of

${ \mathrm{N} }_{ 2 }{ \mathrm{O} }_{ 5 }$ ,

${ \mathrm{p }}_{ 0 }=\dfrac { 2 }{ 5 } \times 584.5=233.8$ mm Hg

After

$30$ minutes, the total pressure

$=284.5$

$mm$

$Hg$

$\begin{matrix} 2{ N }_{ 2 }{ O }_{ 5 } \\ { p }_{ 0 }-2p \end{matrix}\begin{matrix} \longrightarrow \\ \end{matrix}\begin{matrix} 4N{ O }_{ 2 } \\ 4p \end{matrix}\begin{matrix} + \\ \end{matrix}\begin{matrix} { O }_{ 2 } \\ p \end{matrix}$

${ p }_{ 0 }+3p=284.5$

$3p=284.5-233.8=50.7$ mm Hg

$p=\dfrac { 50.7 }{ 3 } =16.9$ mm Hg

Pressure of

${ \mathrm{N} }_{ 2 }{ \mathrm{O} }_{ 5 }$ after

$30$ minutes

$=233.8-\left( 2\times 16.9 \right)$

$=200$ mm Hg

$\mathrm{k}=\dfrac { 2.303 }{ 30 } \log _{ 10 }{ \dfrac { 233.8 }{ 200.0 } }$

$=5.206\times { 10 }^{ -3 }{ \mathrm{min} }^{ -1 }$

Hence, the correct option is

$\text{A}$

The reaction is given below, the rate constant for disappearance of A is

$7.48\times10^{-3}sec^{-1}$ . The time required for the total pressure in a system containing A at an initial pressure of 0.1 atm to rise to 0.145 atm is:

$2A(g)\rightarrow 4B(g)+C(g)$

0%
0.80 min
0%
0.567 min
0%
0.433 min
0%
0.344 min

Explanation

Its a first order reaction and for this,

$kt = 2.303\ log\left( \dfrac{a}{a-x}\right)$

Here,

$2A(g)\rightarrow 4B(g)+C(g)$
P - -
P-2x 4x x

So here, P-2x+4x+x= 0.145
From given condition 2x is 0.03 and

$t = \frac {2.303}{k}\ log\left( \dfrac{p}{p-2x}\right) = 47.63\ seconds = 0.80$ min.

The reaction

$A(aq) \rightarrow B(aq)+C(aq)$ is monitored by measuring optical rotation of reaction mixture at different time interval. The species A, B and C are optically active with specific rotations

$20^{\circ}, 30^{\circ}$ and

$-40^{\circ}$ respectively. Starting with pure A if the value of optical rotation was found to be

$2.5^{\circ}$ after

$6.93^{\circ}$ minutes and optical rotation was

$-5^{\circ}$ after infinite time. Find the rate constant for the first-order conversion into A into B and C:

0%
$0.1 min^{-1}$
0%
$0.2 min^{-1}$
0%
$0.3 min^{-1}$
0%
$0.4 min^{-1}$

Explanation

As after completion of the reaction rotation is

$-5 ^{\circ}$ So according to first order reaction

$k = 2.303log(r_0 - r_{\infty })/(r_t - r_{\infty })/6.93 = 0.1/min$

Time	t	$\infty$
Rotation of Glucose & Fructose	$r_{t}$	$r _{\infty }$

$S\rightarrow$

$G+F$
What id the value of

$k$ for the above reaction under given circumstances?

0%
$\displaystyle k=\frac{1}{t}ln\frac{r_{\infty }}{(r_{\infty }-r_{t})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{r_{t}}{(r_{\infty }-r_{t})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{r_{\infty }}{(r_{t}-r_{\infty})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{r_{t}}{(r_{t}-r_{\infty})}$

Explanation

The rate law expression for the first order reaction is

$k=\frac {2.303} {t}log \frac {a} {a-x}$ .

But

$a \propto r _{\infty }$ and

$x \propto r_{t}$

Substitute values in the above equation.

$\displaystyle k=\frac{1}{t}ln\frac{r_{\infty }}{(r_{\infty }-r_{t})}$

A metal slowly forms an oxide film which completely protects the metal when the film thickness is 3.956 thousandths of an inch. If the film thickness is 1.281 thou. in 6 weeks, then the time it will take before it is 2.481 thou. is :(the rate of film formation follows first-order kinetics)

0%
12 weeks
0%
14 weeks
0%
15 weeks
0%
17 weeks

Explanation

When the film thickness is 3.956 thousandths of an inch.,

$a=\frac {3.956} {1000}$ .
For

$t=6 weeks, a-x=\frac {(3.956-1.281)} {1000}$

$k=\frac {2.303} {t} log \frac {a}{a-x}=\frac {2.303} {6} log \frac {3.956}{3.956-1.281}$ ......(1)
For

$t=? weeks$ ,

$a-x=\frac {(3.956-2.481)} {1000}$

$k=\frac {2.303} {t} log \frac {a}{a-x}=\frac {2.303} {t} log \frac {3.956}{3.956-2.481}$ ......(2)
From equations (1) and (2),

$\frac {2.303} {6} log \frac {3.956}{3.956-1.281}=\frac {2.303} {t} log \frac {3.956}{3.956-2.481}$
Hence,

$t=12 weeks$ .

$100^{\circ}$ C the gaseous reaction

$A\rightarrow$ 2B+C was observed to be of first order. On starting with pure A it is found that at the end of 10 minutes the total pressure of system is 176mm.Hg and after a long time 270mm Hg.

Initial pressure of A is:

0%
90mm
0%
80mm
0%
70mm
0%
60mm

Explanation

After long time the dissociation of A is complete. The dissociation of one mole of A gives 3 moles of products. At this time, total pressure is 270mm Hg. Hence, the initial pressure of

$A=\frac{270}{3}=90mm$ .

For a reversible first - order reaction

$A\rightleftharpoons B K_{1}=10^{-2}s^{-1}$ and

$[B]_{eq}=4$ . If

$[A]_{0}=0.01$ mole

$L^{-1}$ and

$[B]_{0}=0$ , what will be the concentration of B after 30 s?

0%
0.0025 m
0%
0.0013 m
0%
0.0026 m
0%
0.0030 m

Explanation

As we know,

$k_1+k_2 = (1/t)log(x_e - x_0)/(x_e - x)$ and

$k_e = k_1/k_2$
And from given data, conc. of B after 30 sec.

$k_1+k_2 = 1/30)log(4-0/4-x)$

$x= 0.0025$ m

In this case, we have (Consider it as a first-order reaction)

$A\rightarrow B+C$
Time

$t$

$\infty$
Total pressure

$p_{2}$

$\ \ \:\:\:\:p_{3}$
Then

$k$ is:

0%
$\displaystyle k=\frac{1}{t}ln\frac{p_{3}}{2(p_{3}-p_{2})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{p_{2}}{2(p_{3}-p_{2})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{p_{3}}{2(p_{1}-p_{2})}$
0%
$\displaystyle k=\frac{1}{t}ln\frac{p_{1}}{2(p_{3}-p_{2})}$

Explanation

When

$t=\infty$ , entire

$A$ has reacted. 1 mole of

$A$ gives 1 mole of

$B$ and 1 mole of

$C$

$a=\dfrac {p_3} {2}$ .

$t=t$ , the total pressure is

$a-x+x+x=a+x=p_2$ .
Hence,

$x=p_2-a$ .

$a-x=a-(p_2-a)= 2a-p_2=p_3-p_2$
The integrated rate law for the first order reaction is

$k=\dfrac {1} {t} ln{\dfrac {[A]_0} {[A]_t}}=\dfrac {1} {t} ln {\dfrac {P_3} {2}}/( {P_3-P_2})=\displaystyle k=\dfrac{1}{t}ln\dfrac{p_{3}}{2(p_{3}-p_{2})}$

$\displaystyle SO _{2}Cl _{2}(g)\rightarrow SO_{2}(g)$

The given reaction is a first order gas reaction with

$k=2.2 \times \displaystyle 10^{-5}sec^{-1}$ at

$320^{\circ}$ C. What % of

$SO_{2}Cl_{2}$ is decomposed on heating this gas for 90 min?

0%
11.2%
0%
12.3%
0%
13.4%
0%
14.5%

Explanation

for a first order reaction:

$k = 2.303\ log\left(\cfrac{\frac{a}{a-x}}{t}\right);$

$k=2.2\times 10^{-5} = 2.303log\left(\cfrac{\frac{a}{a-x}}{90 \times 60}\right)$

$\cfrac{(a-x)\times100 }a = 88.8$ %

so % decomposed

$= 100-88.8= 11.2$ %

Hence, the correct option is

$\text{A}$

The decomposition of a compound

$P$ , at temperature

$T$ according to the equation

$\displaystyle2P_{(g)}\rightarrow4Q_{(g)}+R_{(g)}+S_{(l)}$ is the first order reaction. After

$30$ minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be

$317$ mm

$Hg$ and after a long period of time the total pressure observed to be

$617$ mm

$Hg$ .

The total pressure of the vessel after

$75$ minute, if the volume of liquid

$S$ is supposed to be negligible is:

(Given : Vapour pressure of

$S (l)$ at temperature

$T=32.5$ mm

$Hg$ )

0%
$P_{t}=379.55$ mm $Hg$
0%
$P_{t}=387.64$ mm $Hg$
0%
$P_{t}=468.23$ mm $Hg$
0%
$P_{t}=489.44$ mm $Hg$

Explanation

After long time, total pressure is 617 mm Hg.

$5P=617-32.5$

$2P=255.4$ .....(1)
After 30 min,

$255+4x=317-32.5$

$x=7.375$ ......(2)
The expression for the rate of the reaction is

$k=\frac {2.303} {t} log {a}{(a-x)}=\frac {2.303} {30} log ({255}\times {(247.62)})$ ......(3)
After 75 min,

$k=\frac {2.303} {t} log {a}{(a-x)}=\frac {2.303} {(75)} log( {255}\times {(255-y)})$ ......(4)
From (3) and (4),

$y=23$
Total pressure

$= 255+4y+32.5=255+4(23)+32.5=379.55\ mm\ Hg$

$A(aq)\longrightarrow B(aq)+C(aq)$ is a first order reaction.

Time	$t$	$\infty$
moles of reagent	${ n }_{ 1 }$	${ n }_{ 2 }$

Reaction progress is measure with help of titration '

$R$ '. If all

$A,B$ and

$C$ reacted with reagent and have '

$n$ ' factors [

$n$ factor;

$eq.wt=\cfrac {mol.wt.}{n}$ ] in the ratio of

$1:2:3$ with the reagent. The

$k$ in terms of

$t,{ n }_{ 1 }$ and

${ n }_{ 2 }$ is :

0%
$k=\cfrac { 1 }{ t } \ln { \left( \cfrac { { n }_{ 2 } }{ { n }_{ 2 }-{ { n }_{ 1 } } } \right) }$
0%
$k=\cfrac { 1 }{ t } \ln { \left( \cfrac { 2{ n }_{ 2 } }{ { n }_{ 2 }-{ { n }_{ 1 } } } \right) }$
0%
$k=\cfrac { 1 }{ t } \ln { \left( \cfrac {4 { n }_{ 2 } }{ { n }_{ 2 }-{ { n }_{ 1 } } } \right) }$
0%
$k=\cfrac { 1 }{ t } \ln { \left( \cfrac {4 { n }_{ 2 } }{5( { n }_{ 2 }-{ { n }_{ 1 }) } } \right) }$

Explanation

The expression for the rate constant for the first order reaction is

$k=\cfrac { 1 }{ t } \ln { \left( \cfrac {[A]_0 }{[A]_t } \right) }$
But

${ \left( \cfrac {[A]_0 }{[A]_t } \right) }={ \left( \cfrac {4 { n }_{ 2 } }{5( { n }_{ 2 }-{ { n }_{ 1 }) } } \right) }$
The expression for the rate constant for the first order reaction becomes

$k=\cfrac { 1 }{ t } \ln { \left( \cfrac {4 { n }_{ 2 } }{5( { n }_{ 2 }-{ { n }_{ 1 }) } } \right) }$

Two first order reaction have half-lives in the ratio

$8:1$ . Calculate the ratio of time intervals

${ t }_{ 1 }$ and

${ t }_{ 2 }$ are the time period for the

${ \left( \cfrac { 1 }{ 4 } \right) }^{ th }$ and

${ \left( \cfrac { 3 }{ 4 } \right) }^{ th }$ completion.

0%
$1:0.301$
0%
$0.125:0.602$
0%
$1:0.602$
0%
none of these

Explanation

As their half-lives in the ratio

$8:1$ so their rate constant ratio is

$1:8$ so ratio of time taken is

$t_1/t_2 = \cfrac { (2.303/k_1)log(1/(1-1/4)) }{(2.303/k_2)log(1/(1-3/4)) } = \cfrac {8 \times log(4/3) }{log4 } = 1 : 0.602$

The gaseous decomposition reaction,

$A(g)\longrightarrow 2B(g)+C(g)$ is observed to first order the excess of liquid water at

${ 25 }^{ o }C$ . It is found that after

$10$ minutes the total pressure of system is

$188$ torr and after very long time it is

$388$ torr. The rate constant of the reaction (in

$hr^{ -1 }$ ) is:
[Given: vapour pressure of

${ H }_{ 2 }O$ at

${ 25 }^{ o }C$ is

$28$ torr. (

$\ln { 2 } =0.7,\ln { 3 } =1.1,\ln { 10 } =2.3$ )]

0%
$0.02$
0%
$1.2$
0%
$0.2$
0%
none of these

Explanation

By putting values in equation

$k=\dfrac{2.303}{t}log\dfrac { P_0 }{ (P_0 - x) } = \dfrac{2.303}{t}log\dfrac { P_\infty/3 }{ (P_\infty /3 - (P_t - P_0)) }$

We will get answer as

$k=1.2 hr^{-1}$

$(A, B, C)$ let the initial pressure of

$A$ be

$P_s$ mm of

$Hg$ and the pressure of

$A$ decreases in 10 minute be

$x$ unit.

$\begin{matrix} & & \\ &A(g)&\longrightarrow &2B(g)&+&C(g) \\ \text{Initially} & P_0 &&0&&0\\\text{At time 10 min} &P_0-x & &2x& &x \\\text{At}\, \infty \,\text{time} & 0 &&2P_0& &P_0\end{matrix}$

After long time interval,

$P_{\infty} = 2P_0 + P_0 = 3P_0$

After

$t$ time

$P_t = P_0 + 2x$

For a first order reaction, the rate constant expression would be

$k=\dfrac{2.303}{t}\log \dfrac{P_0}{P_0-x}$

A compound

$A$ dissociate by two parallel first order paths at certain temperature

$A(g)\xrightarrow [ ]{ { k }_{ 1 }({ min }^{ -1 }) } 2B(g)$

${ k }_{ 1 }=6.93\times { 10 }^{ -3 }min^{ -1 }$

$A(g)\xrightarrow [ ]{ { k }_{ 2 }({ min }^{ -1 }) } C(g)$

${ k }_{ 2 }=6.93\times { 10 }^{ -3 }min^{ -1 }$
The reaction started with

$1$ mole of pure '

$A$ ' in

$1$ litre closed container with initial pressure

$2$ atm. What is the pressure (in atm) developed in container after

$50$ minutes from start of experiment?

0%
$1.25$
0%
$0.75$
0%
$1.50$
0%
$2.50$

Explanation

For both reactions, half life is 100 min so after 50 min no. of moles of all reagents in solution is 1.25 mole. So the pressure of the container is

$2 \times 1.25/1 = 2.50$ atm.

For a first-order homogeneous gaseous reaction,

$A\longrightarrow 2B+C$ .
If the total pressure after time

$t$ was

${ P }_{ t }$ and after long time

$(t\rightarrow \infty )$ was

${ P }_{ \infty }$ then

$k$ in terms of

${ P }_{ t },{ P }_{ \infty }$ and

$t$ is :

0%
$k=\cfrac { 2.303 }{ t } \log { \left( \cfrac { 2{ P }_{ \infty } }{ { P }_{ \infty }-{ P }_{ t } } \right) }$
0%
$k=\cfrac { 2.303 }{ t } \log { \left( \cfrac { 2{ P }_{ \infty } }{2( { P }_{ \infty }-{ P }_{ t }) } \right) }$
0%
$k=\cfrac { 2.303 }{ t } \log { \left( \cfrac {2 { P }_{ \infty } }{3( { P }_{ \infty }-{ P }_{ t }) } \right) }$
0%
none of these

Explanation

$A \rightarrow 2B + C$

$t = 0$

$P_{i}$

$0$

$t$

$P_{t} - x$

$2x$

$x$

$t \rightarrow \infty$

$0$

$2P_{i}$

$P_{i}$

$P_{\infty} = 3P_{i}$ or

$P_{i} = \dfrac{P_{\infty}}{3}; P_{i} + 2x = P_{t}$

$x = \dfrac{P_t - P_i}{2}$

As we know

$k = \dfrac{2.303}{t} log \left ( \dfrac{P_i}{P_i - x} \right )$

$k = \dfrac{2.303}{t} log \left ( \dfrac{2P_{\infty}}{3(P_{\infty} - P_t)} \right )$

Hence optoin (C) is correct.

The reaction

$A(g)\longrightarrow B(g)+2C(g)$ is a first order reaction with rate constant

$2.772\times { 10 }^{ -3 }s^{ -1 }$ . Starting with

$0.1$ mole of

$A$ in

$2$ litre vessel, find the concentration of

$A$ after

$250sec$ when the reaction is allowed to take place at constant pressure at

$300K$ ? (Given ln2= 0.693)

0%
$0.0125M$
0%
$0.025M$
0%
$0.05M$
0%
none of these

Explanation

Half life of reaction is

$t_{1/2} = 0.693/k = 250$ sec.
So after 250 sec concentration become half and remaing concentration is

$0.05/2 = 0.025M$ .

$\displaystyle \:A+B\rightleftharpoons AB+I\xrightarrow{k_{2}}P+A$

$k_1$ is the rate constant of the reversible step and If

$k_1$ is much smaller than

$k_{2}$ , The most suitable qualitative plot of potential energy (P.E.) versus reaction coordinate for the above reaction.

Assertion: In a reversible endothermic reaction,

$E_{act}$ of forward reaction is higher than that of backward reaction.

Reason: The threshold energy of forward reaction is more than that of backward reaction.

0%
Both Assertion and Reason are true and Reason is the correct explanation of Assertion
0%
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
0%
Assertion is true but Reason is false
0%
Assertion is false but Reason is true
0%
Both Assertion and Reason are false

The reaction

$cis-X\overset { { k }_{ f } }{ \underset { { k }_{ b } }{ \rightleftharpoons } } tran-X$ is first order in both directions. At

${ 25 }^{ o }C$ , the equilibrium constant is

$0.10$ and the rate constant

${ k }_{ f }=3\times { 10 }^{ -4 }s^{ -1 }$ . In an experiment starting with the pure

$cis-$ form, how long would it take for half of the equilibrium amount of the

$trans-$ isomer to be formed?

0%
$150$ sec
0%
$200$ sec
0%
$155$ sec
0%
$210$ sec

For a given reaction of the first order, it takes 15 minutes for the concentration to drop from 0.8 M to 0.4 M. The time required for the concentration to drop from 0.1 M to 0.025 M will be :

0%
15 minutes
0%
60 minutes
0%
30 minutes
0%
7.5 minutes

Explanation

The reactant concentration drop from

$0.8\ M$ to

$0.4\ M$ , i.e.,

$50\%$ takes place in 15 minutes.

$K=\dfrac {2.303}{15} log \dfrac {0.8}{0.4}=\dfrac {0.693}{15}=0.0462 \ min^{-1}$

Also,

$t=\dfrac {2.303}{K}log \dfrac {0.1}{0.025}$

$=\dfrac {2.303}{0.0462} log \dfrac {0.1}{0.025}=30\ min$

The study of chemical kinetics becomes highly complicated if there occurs:

0%
reversible reaction
0%
side reaction
0%
surface reaction
0%
none of these

A reaction occurs in parallel paths. For each path having energy of activation as

$E,\ 2E,\ 3E,\ ... nE$ and rate constant

$K,\ 2K,\ 3K,\ .... nK$ respectively. If

$E_{AV}=3E$ , then find out the value of

$n$ .

0%
$4$
0%
$5$
0%
$6$
0%
None of these

Explanation

The average energy is given by the expression

$\displaystyle 3E=\frac {E+2E+3E+...+nE}{n}=\frac {n(n+1)}{2n}E$

Hence,

$\displaystyle\frac {n+1}{2}=3$

$6=n+1$

$n=5$

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 10 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 10 - MCQExams.com

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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