Thermodynamics - Class 11 Engineering Chemistry - Extra Questions

Total internal energy of a system can never be evaluated but the change in internal energy can be determined by 1st law of thermodynamics.
If true enter 1, else enter 0.

The law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but cannot be created or destroyed. 

Total internal energy of a system can never be evaluated but the change in internal energy can be determined by 1st law of thermodynamics. 

The first law is often formulated by stating that the change in the internal energy of a closed system is equal to the amount of heat supplied to the system, minus the amount of work done by the system on its surroundings.

Thus the statement is true. (1)


Write mathematical equations of first law of thermodynamics for the following processes :
(a) Adiabatic process                 (b) Isochoric process

(a) In adiabatic process, there is no exchange of heat thus $$q=0$$. So First law becomes $$\Delta U=W$$.

(b) Isochoric process, as it is carried in closed vessel $$\Delta V=0$$ thus $$W=0$$. So first law becomes $$\Delta U=q$$.


What is Gibb's Energy?

Gibb's Energy: A thermodynamic quantity equal to the enthalpy (of a system or process) minus the product of the entropy and the absolute temperature

It is a state function defined by the following expression.

$$\displaystyle G=H-TS$$

Here, G is Gibb's Energy, H is enthalpy, S is entropy and  T is temperature.

The change in Gibb's Energy is given by the following expression.
$$\displaystyle \Delta G = \Delta H - T \Delta  S$$




The thermal capacity of calorimeter system is $$17.7\ kJ$$ $${K}^{-1}$$. ($$R=8.313\ {mol}^{-1}{K}^{-1}$$) (only magnitude in nearest integer in kJ/mol)

At constant volume 
$$\Delta H = \Delta U$$ 

Rise in temperature of the system = $$0.5^{o} C= 0.5 K$$ 

Total heat released $$= -C_{p}\Delta T = -17.7\frac{kJ}{K}\times 0.5 K=-8.85\; kJ$$

0.16 g of methane $$=\frac{0.16}{16}$$ mole$$ =0.01$$ mole of methane  (molecular weight of methane =16 g)

Heat of combustion $$=\frac{tota\; heat\; released}{number\; of\; moles} = \frac{-8.85}{0.01} = -885 \; kJmol^{-1}$$

Answer = 885



For a given reaction, energy of activation for forward reaction $$(E_{af}) \: is \: 80kJ.mol^{-1} .\Delta H =-40kJ.mol^{-1}$$  for the reaction. A catalyst lowers $$E_{af} \: to \: 20 \: kJ.mol^{-1}$$ . Find out the ratio of energy of activation for reverse reaction before and after addition of catalyst.

$$\Delta H=E_f- E_b        -40= 80 -E_b  ,     \ \ \        E_b=120 \: kJ/mole$$

Catalyst lower the $$E_{af} \:$$ to $$\: 20 \:kJ/mole$$ for forward Rxn then $$E'_f =20kJ/mol$$

We know catalyst decreases the activation energy equal amount in both directions:

$${E_{b}}^{'}=(120 - 60) = 60 \:kj/mol$$

$$\cfrac{E_b}{E'_b}=\cfrac{120}{60}=2.0$$


First law of thermodynamics is not adequate in predicting the spontaneity of process.
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True,
For predicting the spontaneity of the process we need bothe the enthalpy and the entropy data. The change in the entropy is predicted by the second law of thermodynamics. 
Hence, the given statement is correct


What is the relationship between $$\Delta G, \ \Delta H \ \& \Delta S$$?

Relationship between  $$\Delta G, \ \Delta H \ \& \Delta $$ is given by:
$$\Delta G= \Delta H -T \Delta S $$


Calculate $$\Delta { G }^{ o }$$ for the following reaction:
$$CO(g)+\cfrac { 1 }{ 2 } { O }_{ 2 }(g)\longrightarrow { CO }_{ 2 }(g);\Delta { H }^{ o }=-282.84kJ\quad $$
Given, $${ S }_{ { CO }_{ 2 } }^{ o }=213.8J{ K }^{ -1 }{ mol }^{ -1 },{ S }_{ CO(g) }^{ o }=197.9J{ K }^{ -1 }{ mol }^{ -1 },{ S }_{ { O }_{ 2 } }^{ o }=205.0J{ K }^{ -1 }{ mol }^{ -1 }$$

$$\Delta { S }^{ o }=\sum { { S }_{ (products) }^{ o } } -\sum { { S }_{ reactants }^{ o } } $$
$$=\left[ { S }_{ { CO }_{ 2 } }^{ o } \right] -\left[ { S }_{ CO(g) }^{ o }+\cfrac { 1 }{ 2 } { S }_{ { O }_{ 2 } }^{ o } \right] $$
$$=213-[197.9+\cfrac { 1 }{ 2 } \times 205]=-86.6J{ K }^{ -1 }\quad $$
According to Gibbs-Helmholtz equation
$$\Delta { G }^{ o }=\Delta { H }^{ o }-T\Delta { S }^{ o }=-282.84-298\times (-86.6\times { 10 }^{ -3 })$$
$$=-257.033kJ$$


The standard enthalpy and entropy changes for the reaction in equilibrium for the forward direction are given below:
$$CO(g)+{ H }_{ 2 }O(g)\rightleftharpoons C{ O }_{ 2 }(g)+{ H }_{ 2 }(g)$$
$$\Delta { H }_{ 300K }^{ o }=-41.16kJ{ mol }^{ -1 }$$
$$\Delta { S }_{ 300K }^{ o }=-4.24\times { 10 }^{ -2 }kJ{ mol }^{ -1 }$$
$$\Delta { H }_{ 1200K }^{ o }=-32.93kJ{ mol }^{ -1 }$$
$$\Delta { S }_{ 1200K }^{ o }=-2.96\times { 10 }^{ -2 }kJ{ mol }^{ -1 }\quad $$
Calculate $${K}_{P}$$ at each temperature and predict the direction of reaction at $$300K$$ and $$1200K$$, when $${ P }_{ CO }={ P }_{ { CO }_{ 2 } }={ P }_{ { H }_{ 2 } }={ P }_{ { H }_{ 2 }O }=1$$ atm at initial state.

At $$300K$$
$$\Delta { G }^{ o }=\Delta { H }^{ o }-T\Delta { S }^{ o }$$
$$=41.16-300\times (-4.24\times { 10 }^{ -2 })=-28.44kJ$$
since, $$\Delta { G }^{ o }$$ is negative hence reaction is spontaneous in forward direction.
$$\Delta { G }^{ o }=-2.303RT\log { { K }_{ P } } $$
$$-28.44=-2.303\times 8.314\times { 10 }^{ -3 }\times 300\log { { K }_{ P } } $$
$${ K }_{ P }=893\times { 10 }^{ 4 }$$
At $$1200K$$
$$\Delta { G }^{ o }=\Delta { H }^{ o }-T\Delta { S }^{ o }$$
$$2.59=-2.303\times 8.314\times { 10 }^{ -3 }\times 1200\log { { K }_{ P } } $$
$${K}_{P}=0.77$$


$${ C }_{ 2 }{ H }_{ 4 }+{ Cl }_{ 2 }\longrightarrow { C }_{ 2 }{ H }_{ 4 }{ Cl }_{ 2 }$$
$$\Delta H=-270.6kJ\quad { mol }^{ -1 };\Delta S=-139J{ K }^{ -1 }$$
(i) Is the reaction favoured by entropy, enthalpy both or none?
(ii) Find $$\Delta G$$ if $$T=300K$$

(i) Since, $$\Delta H=-ve$$, exothermic process and is favoured i.e., it will be spontaneous
(ii) $$\Delta G=\Delta H-T\Delta S=-270.6\times 1000-300\times (-139)=-228kJ\quad $$


In a process 701 J of heat is absorbed by a system and 394 J of work is done by system. What is the change in internal energy for the process?

According to 1st law of thermodynamic 
$$\triangle U=q+w$$
Heat absorbed by system $$\Rightarrow q=+701Joule$$
$$w=$$Work done by system
$$w=-394 J$$
$$\triangle U=+701-394$$
$$\triangle U=307 Joule$$


How much energy is required to change $$2\ Kg$$ of ice at $${0}^{0}C$$ into water at $${20}^{0}C$$? (specific latent of heat of fusion of water = $$3,34,000J/Kg$$, specific heat capacity of water = $${4200}\ {J}{Kg}^{-1}{K}^{-1}$$).

Total heat= $$\left[\text{Heat required to convert 2 kg of ice to 2 kg of water at } O^oC\right]+$$
                    $$\left[\text{Heat required to convert 2 kg of water at } o^oC\text{ to 2kg of water at}20^oC\right]$$
$$=mh_{fg}+mC_p\Delta T$$
$$=(2 \times 334 kJ)+(2 \times 4.2 \times 20)$$
$$=668+168=836 kJ$$ of heat is required.


State and explain the first law of thermochemistry.

If enthalpy of formation of a compound is $$-xkJ$$, enthalpy of its decomposition in to its elements is $$+xkJ$$.

Example: $$C(s)+O_{2}(g)\rightarrow CO_{2}$$;

$$\Delta H=-xkJ$$

According to the law (Lavoisier & Laplace law) $$CO_{2}(g)\rightarrow C(s)+O_{2}(g)$$;

$$\Delta H=+xkJ$$.


Law of conservation of energy is also known as ........

$$\text{The first law of thermodynamics, also known as Law of Conservation of Energy, states that energy can neither }$$ $$\text{be created nor destroyed; energy can only be transferred or changed from one form to another.}$$

Thus, the correct answer is First law of thermodynamics.


List I and List II contains four entries each. Entries of Column I are to be matched some entries of List II. One or more than one entries of List I may match with the same entry of List II.
List I lists the partial derivatives and List II lists the thermodynamic variable.

As, $$dG = VdP - SdT$$ so:

$${ \left( \cfrac { \partial G }{ \partial T } \right) }_{ P } = -S$$ and 

$${ \left( \cfrac { \partial G }{ \partial P } \right) }_{ T } = V$$

$${ \left( \cfrac { \partial U }{ \partial T } \right) }_{ V } = C_v$$

$${ \left( \cfrac { \partial H }{ \partial T } \right) }_{ P } = C_p$$


A slice of banana weighing 2.502 g was burnt in a bomb calorimeter producing a temperature rise of $$3.05^{\circ}C$$. The combustion of 0.316 g of benzoic acid in the same calorimeter produced a temperature rise of $$3.24^{\circ}C$$. The heat of combustion of benzoic acid at constant volume is $$-3227 \: kJ \: mol^{-1}$$. If the average banana mass is 125 g, the kilojoules  of energy can be obtained from 1 average banana is: (nearest integer value)

$$q = mS. \Delta T$$ for combustion of benzoic acid
 ($$Note:$$ In bomb calorimeter reaction is made at constant volume)

The molar mass of benzoic acid= 122g

122 g of benzoic acid produces -3227 $$KJ\ mol^{-1}$$ heat of combustion


                    $$q=\displaystyle \frac{0.316\times 3227}{122}=m\times S\times 3.24$$

$$\therefore $$ Water equivalent of bomb calorimeter $$(mS)$$

                 $$m \times S=\displaystyle \frac{0.316\times 3227}{122 \times 3.24}=2.58 \:kJ/^{\circ}\! C$$

For banana slice; $$q = mS\times \Delta T$$

                           $$q=2.58\times 3.05$$

                              $$=7.87 \: kJ \: per \: 2.502 \: g \: banana$$

$$\displaystyle \therefore Heat \: produced \: by \: 125 \: g \: banana =\frac{7.87\times 125}{2.502}=393.18\: kJ \approx 393 kJ$$

Hence, the heat produced by 125g of banana will be 393 $$kJ$$


Calculate the enthalpy change for the reaction (only magnitude in nearest integer in kj/mol):
$$2C(s)+2{H}_{2}(g)+{O}_{2}(g)\longrightarrow {CH}_{3}{CO}_{2}H(l)$$

Given reaction
$$R: 2C+2H_2+O_2 \rightarrow CH_3CO_2H;\quad \Delta H = ? $$

From the given table, enthalpies of the following reactions are

$$R_1: C+O_2 \rightarrow CO_2;\quad \Delta H_1 = -394\; kJ/mol$$

$$R_2: H_2+\frac{1}{2}O_2 \rightarrow H_2O;\quad \Delta H_2 = -286\; kJ/mol$$

$$R_3: CH_3CO_2H+2O_2 \rightarrow 2CO_2+2H_2O;\quad \Delta H_3 = -876\; kJ/mol$$

And

$$R= -R_3 + 2R_1 + 2R_2$$

$$\Rightarrow \Delta H = -\Delta H_3+2\Delta H_1+2\Delta H_2 =+876+2\times (- 394) +2\times (-286)  = -484\; kJ/mol $$

Answer = 484


What are the characteristics of free energy $$(G)$$?

i) $$G$$ is defined as $$H-TS$$ where $$H$$ and $$S$$ are enthalpy and entropy of the system respectively $$T=$$ temperature. Since $$H$$ and $$S$$ are state functions, $$G$$ is a state function.
ii) $$G$$ is an extensive property while $$\Delta G=G_{2}-G_{1}$$ which is the free energy change between the initial and final states of the system.
iii) $$G$$ has a single value for the thermodynamic state of the system.
iv) $$G$$ and $$\Delta G$$ value correspond to the system only.
There are three cases of $$\Delta G$$ in predicting the nature of the process.
a) When $$\Delta G<0$$, (Negative), the process is spontaneous and feasible.
b) When $$\Delta G=0$$, the process is in equilibrium.
c) When $$\Delta G> 0$$, (Positive), the process is non-spontaneous and not feasible.
v) $$\Delta G=\Delta H-T\Delta S$$ but according of $$I$$ law of thermodynamics.
$$\Delta H=\Delta E+P\Delta V$$ and $$\Delta E = q-w$$
$$\therefore \Delta G=q-w+P\Delta V-T\Delta S$$
but $$\Delta S-q/T$$ and $$T\Delta S=q=$$ heat involved in the process.
$$\therefore \Delta G=q-w+P\Delta V-q=w+P\Delta V$$
(or) $$-\Delta G=w-P\Delta V=$$ Network.
The decrease in free energy $$-\Delta G$$, accompanying a process taking place at constant temperature and pressure is equal to maximum obtainable work.
Network $$=-\Delta G=w-P\Delta V$$


The enthalpy of combustion of glucose is $$-2808kJ\quad { mol }^{ -1 }$$ at $${ 25 }^{ o }C$$. How many grams do you need to consume. (Assume $$wt=62.5kg$$)
(a) to climb a flight of stairs rising through $$3M$$
(b) to climb a mountain of altitude $$3000M$$?
Assume $$25$$% of enthalpy can be converted to useful work.

Combustion reaction of glucose is:
$$C_6H_{ 12 }O_6+6O_2\rightarrow 6CO_2+6H_2O$$; $$\Delta H_{ comb }=-2808\ kJ/mol$$
Molecular wt of glucose is $$180.156\ g/mol$$
(a) Potential energy required to climb 3 m is:
$$E=m.g.h$$
where $$m=62.5\ kg$$, $$g$$=acceleration of gravity=$$9.8\ m/s^2$$ and $$h=3\ m$$
Upon substitution we get:
$$E=62.5\times 9.8\times 3$$= $$1837.5\ J$$
Since only 25% of enthalpy of combustion can be converted to useful work
therefore amount of energy required is:
$$x\times 0.25=1837.5\ J$$
$$x=7350\ J=7.35\ kJ$$
Since 1 mol ($$180.15\ g$$) glucose provides $$2808\ kJ$$ energy
Therefore $$7.35\ kJ$$ energy will be obtained by $$\frac { 180.15\times 7.35 }{2808}g$$ Glucose
=$$0.47\ g$$ Glucose.
(b) As for climbing $$3\ m$$ $$0.47\ g$$ glucose is required, therefore to climb $$3000\ m$$ amount of glucose required will be= $$0.47\times { \frac {3000}{3} }=0.47\ kg$$


A sample of $$0.16\ g\ CH_{4}$$ was subjected to the combustion at $$27^{o}C$$ in a bomb calorimeter. The temperature of the calorimeter. The temperature of the calorimeter system (including water) was found to rise by $$0.5^{o}C$$. Calculate the heat of combustion of methane at
(i) constant volume and
(ii) constant pressure. the thermal capacity of calorimeter system is $$17.0\ kJ\ K^{-1}$$ and $$R=8.314\ J\ K^{-1}\ mol^{-1}$$

Amount of heat required = $$0.5$$ degrees =$$17.7$$x$$ 0.5$$
                                                              =$$8.85KJ$$
0.16g of methane produces 8.85KJ
Heat liberated by 1 mol of methane =$$\frac { 8.85 }{ 0.16 } \times 16=885KJ$$
Change in heat =change in energy+nRT
Change of $$n=2$$
$$\Delta H$$=$$ -885+(-2)(8.3\times 1{ 0 }^{ -3 })$$
      = $$-889.986KJ/Mol$$           



The specific heats of iodine vapour and solid are $$0.031$$ and $$0.05$$ cals/g respectively. If heat of sublimation of iodine is $$24$$ cals/g at $${ 200 }^{  o }C$$.
Calculate its value at  $${ 250 }^{ o }C$$.

$${C}_{P} =$$ molar heat capacity of product $$-$$ molar heat capacity of reactant
Molar heat capacity $$=$$ specific heat capacity $$\times$$ molecular weight
Molecular weight of iodine $$= 127 \; g$$
Given that the specific heats of iodine vapour and solid are $$0.031$$ and $$0.05 \; {cal}/{g}$$ respectively.
Molar heat capacity of iodine solid $$= 0.05 \times 127 = 6.35$$
Molar heat capacity of iodine vapour $$= 0.031 \times 127 = 3.9$$
$$\therefore {C}_{P} = 6.35 - 3.9 = 2.45 \; {cal}/{g}$$
According to Kirchoff's equation-
$${C}_{p} = \cfrac{{\Delta{H}}_{2} - {\Delta{H}}_{1}}{{T}_{2} - {T}_{1}}$$
Where as,
$${\Delta{H}}_{1} =$$ Heat of sublimation at temperature $${T}_{1}$$
$${\Delta{H}}_{2} =$$ Heat of sublimation at temperature $${T}_{2}$$
$$\therefore 2.45 = \cfrac{{\Delta{H}}_{2} - 24}{250 - 200}$$
$$\Rightarrow {\Delta{H}}_{2} - 24 = 50 \times 2.45$$
$$\Rightarrow {\Delta{H}}_{2} = 122.5 + 24 = 146.5 \; {cal}/{g}$$
Hence the required answer is $$146.5 \; {cal}/{g}$$.


If 1.0 k cal of heat is added to 1.2 L of $$O_2$$ in a cylinder of constant pressure of 1 atm, the volume increases to 1.5 L. Calculate $$\Delta H$$ and $$\Delta U$$ of the process.

$$\Delta U= q- P \Delta v$$
$$\Rightarrow \Delta U= 1000 - [1\times (1.5- 1.2)]\times 24.21.1.Latm=24.21 cal$$
$$\Rightarrow \Delta U= 999.7 cal$$
or
$$\Rightarrow \Delta U =99.3 Kcal$$
at constant pressure
$$\Delta H= q_p=1Kcal$$


Give reasons : 
Thermodynamically an exothermic reaction is sometimes not spontaneous.

If $$\Delta S$$ is largely (-ve), $$\Delta G = +ve$$
hence, non spontaneous


For the synthesis of ammonia at 300 K:$$N_2(g) +3H_2(g) \rightarrow 2NH_3(g)$$ 
Calculate the value of the $$\Delta G^0$$ in Kcal and give your answer in magnitude by using the following data:
                                                       $$N_2$$            $$H_2$$      $$NH_3$$ 
$$\Delta H^0_{f} (kcal/mole)$$                       0                0                   -10
$$S^0(Cal/K-mole)$$                      40                   30                 45


$$\Delta H^0_R=2\times \Delta H^0_{f(NH_3)}-3\times \Delta H^0_{f(H_2)}-\Delta H^0_{f(N_2)}=-20kcal$$

$$\Delta S^0_R=2\times S^0_{(NH_3)}-3\times S^0_{(H_2)}-S^0_{(N_2)}=2\times 45-3\times 30-40=-40Cal/K$$

$$\Delta G^0=\Delta H^0-T\Delta S^0=-20- \dfrac{300 \times (-40)}{1000}=-20+12=-8Kcal$$

Magnitude of $$\Delta G=8 kcal$$


The heat of solution of $$NH_{4}NO_{3}$$ in water was determined by measuring the amount of electrical work needed to compensate for the cooling which would otherwise occur when the salt dissolves. After the $$NH_{4}NO_{3}$$ was added to the water, electrical energy was provided by the passage of current through a resistance coil until the temperature of the solution reached the value it had prior to the addition of salt. In a typical experiment, $$4.4\ g$$ of $$NH_{4}NO_{3}$$ was added to $$200\ g$$ water. A current of $$0.75$$ ampere was provided through the heater coil, and the voltage across the terminals was $$6.0\ V$$. The current was applied for $$5.2$$ minute. Calculate $$\Delta H$$ for the solution of $$1.0$$ mole $$NH_{4}NO_{3}$$ in enough water to give the same concentration as was attained in the above experiment.


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Given below are some standard heats of reaction;
(a) Heat of formation of water $$=-68.3\ kcal$$
(b) Heat of combustion of acetylene $$=-310.6\ kcal$$
(c) Heat of combustion of ethylene $$=-337.2\ kcal$$
Calculate the heat of reaction for the hydrogenation of acetylene at constant volume at $$25^oC.$$


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Calculate pH at which the following conversion (reaction) will be at equilibrium in basic medium $$I_2(s) \rightarrow I^-(aq)+IO_3^-(aq)$$ when the equilibrium concentrations at 300K are $$ [I^-]=0.10 M$$ and $$ [IO_3^-]=0.10 M$$ 
[Given that $$\Delta G ^0_f(I^-,aq)=-50 kJ/mole, \Delta G^0_f(IO_3^-,aq)=-123.5 kJ/mole, \Delta G^0_f(H_2O,l)=-233kJ/mole$$
$$\Delta G^0_f(OH^-,aq)=-150kJ/mole$$ ideal gas constant=$$R=\dfrac{25}{3}Jmole^{-1}K^{-1}, log_e10=2.3$$]


Balanced equation will be:
$$3I_2(s)+6OH^-\rightleftharpoons 5I^-(aq)+IO_3^-(aq)+3H_2O(l)$$

$$\Delta G^0=5\times (-50)+(-123.5)+3\times (-233)-0-6\times (-150)$$

$$\Delta G^0=-172.5 kJmole^{-1}=-\dfrac{25}{3}\times 300\times 2.3\times 10^{-3}logk$$

$$log k=30$$

$$10^{30}=\dfrac{10^{-5}\times 10^{-1}}{[OH^-]^6}$$

$$\because [OH^-]=10^{-6}$$

$$\because pH=14-6=8$$


$$\alpha- D$$ glucose undergoes muta rotation to $$β-D$$ glucose in aqueous solution. If at $$298 K$$ there is $$60\%$$ conversion. Calculate $$\Delta G^o$$ for the reaction.


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What is specific heat capacity?

It is defined as the quantity of heat required to raise temperature of $$1$$ gram of a substance by $$1^{0}C$$ or $$1K$$.


Class 11 Engineering Chemistry Extra Questions