C (graphite) + O 2 ( g ) → C O 2 ( g ) ; Δ H = − 94 .05 k c a l m o l − 1 C (diamond) + O 2 ( g ) → C O 2 ( g ) ; Δ H = − 94 .50 k c a l m o l − 1 which of the following is correct?

Graphite is the stabler allotrope

C (graphite) → C (diamond) ; Δ H 298 K o = − 450 c a l m o l − 1

C (diamond) → C (graphite) ; Δ H 298 K o = + 450 c a l m o l − 1

Diamond and graphite both are sp 2 hybridised

For which change ∆ H ≠ ∆ E

H C l + N a O H → N a C l + H 2 O

Chemistry - Thermodynamics - Quiz-6

Which of the following compounds will absorb the maximum quantity of heat when dissolved in the same amount of water ? The heats of solution of these compounds at 25°C in kJ/mole of each solute is given in brackets [AMU (Engg.) 2000]

0%

$H N O_{3} (Δ H = - 33)$
0%

$K C l (Δ H = + 17 .64)$
0%

$N H_{4} N O_{3} (Δ H = + 25 .5)$
0%

$H C l (Δ H = - 74 .1)$

A system is changed from state A to state B by one path and from B to A by another path. If E₁ and E₂ are the corresponding changes in internal energy, then [Pb. PMT 2001]

0%

$E_{1} + E_{2} = - v e$
0%

$E_{1} + E_{2} = + v e$
0%

$E_{1} + E_{2} = 0$
0%

None of the above

If (i) $C + O_{2} \to C O_{2}$ , (ii) $C + 1 / 2 O_{2} \to C O$ , (iii) $C O + 1 / 2 O_{2} \to C O_{2}$ , the heats of reaction are Q, –12, –10 respectively. Then Q = [Orissa JEE 2004]

0%

– 2
0%

2
0%

– 22
0%

– 16

Adsorption of gases on a solid surface is generally exothermic because [IIT JEE (Screening) 2004]

0%

Enthalpy is positive
0%

Entropy decreases
0%

Entropy increases
0%

Free energy increase

Two mole of an ideal gas is expanded isothermally and reversibly from 1 litre ot 10 litre at 300 K. The enthalpy change (in kJ) for the process is [IIT JEE (Screening) 2004]

0%

11.4 kJ
0%

–11.4 kJ
0%

0 kJ
0%

4.8 kJ

When a gas undergoes adiabatic expansion, it gets cooled due to -

0%

Loss of energy
0%

Fall in pressure
0%

Decrease in velocity
0%

Increase in energy with work done

In an isobaric process, the ratio of heat supplied to the system (dQ) and work done by the system (dW) for diatomic gas is [AFMC 2002]

0%

1 : 1
0%

7 : 2
0%

7 : 5
0%

5 : 7

If for a given substance melting point is T_B and freezing point is T_A, then correct variation shown by graph between entropy change and temperature is [DCE 2001]

The amount of heat measured for a reaction in a bomb calorimeter is

0%

ΔG
0%

ΔH
0%

ΔE
0%

PΔV

A Beckmann thermometer is used to measure -

0%

High temperature
0%

Low temperature
0%

Normal temperature
0%

All temperatures

The heat liberated when 1.89 g of benzoic acid is burnt in a bomb calorimeter at 25°C increases the temperature of 18.94 kg of water by 0.632°C. If the specific heat of water at 25°C is 0.998 cal/g-deg, the value of the heat combustion of benzoic acid is

0%

(1) 771 kcal
0%

(2) 871.2 kcal
0%

(3) 881.1 kcal
0%

(4) 981.1 kcal

$C_{(graphite)} + O_{2} (g) \to C O_{2} (g)$ ; $Δ H = - 94 .05 k c a l m o l^{- 1}$

$C_{(diamond)} + O_{2} (g) \to C O_{2} (g);$ $Δ H = - 94 .50 k c a l m o l^{- 1}$

which of the following is correct?

0%

$C_{(graphite)} \to C_{(diamond)}$ ; $Δ H_{298 K}^{o} = - 450 c a l m o l^{- 1}$
0%

$C_{(diamond)} \to C_{(graphite)};$ $Δ H_{298 K}^{o} = + 450 c a l m o l^{- 1}$
0%

Graphite is the stabler allotrope
0%

Diamond and graphite both are sp² hybridised

Which of the following is not a correct statement? [ (Engg.) 2002]

0%

When ΔG is negative, the process is spontaneous
0%

When ΔG is zero, the process is in a state of equilibrium
0%

When ΔG is positive, the process is non-spontaneous
0%

None of these

Among mass, volume, density and specific volume of a gas, the intensive properties are

0%

Density and specific volume
0%

Volume and density
0%

Specific volume and mass
0%

Density only

The heat of formation of HCl(g) from the reaction

H₂(g) + Cl₂(g) $\to 2 HCl (g)$ ; $∆ H = - 44 k c a l$ is

0%

+44 kcal
0%

-44 kcal
0%

+22 kcal
0%

-22 kcal

If a chemical change is brought about by one or more methods in one or more steps, then the amount of heat absorbed or evolved during the complete course of reaction is same, which ever method was followed. This law is known as-

0%

Le Chatelier's principle
0%

Hess's law
0%

Joule Thomson effect
0%

Trouton's law

If 1.00 kcal of heat is added to 1.2 L of oxygen in a cylinder at constant pressure of 1.000 atm, the volume increases to 1.5 L, Hence $∆ E$ for this process is:

0%

0.993 kcal
0%

1.0073 kcal
0%

0.0993 kcal
0%

1.00073 kcal

Heats of combustion of $C H_{4}, C_{2} H_{4}, C_{2} H_{6}$ are -890, -1411 and -1560 kJ/mole respectively. Which has the lowest fuel value in kJ/gm ?

0%

$C H_{4}$
0%

$C_{2} H_{4}$
0%

$C_{2} H_{6}$
0%

All same

The standard heat of combustion of a hydrocarbon compound is an/a-

0%

Extensive property
0%

Colligative property
0%

Intensive property
0%

Constitutive property

The temperature of 5 ml of a strong acid increases by $5^{\circ}$ when 5 ml of a strong base is added to it. If 10 ml of each is mixed, temperature should increase by-

0%

$5^{\circ}$
0%

$10^{\circ}$
0%

$15^{\circ}$
0%

cannot be known

The heat of neutralisation of HCl by NaOH is -55.9 kJ/mole. If the heat of neutralisation of HCN by NaOH is -12.1 kJ/mole, then energy of dissociation of HCN is-

0%

-43.8 kJ
0%

43.8 kJ
0%

68 kJ
0%

-68 kJ

For a reaction at equilibrium-

0%

$∆ G = ∆ G^{\circ} = 0$
0%

$∆ G = 0 b u t n o t ∆ G^{\circ}$
0%

$∆ G^{\circ} = 0 b u t n o t ∆ G$
0%

$∆ G = ∆ G^{\circ} \neq 0$

Using only the following data:

(I) $F e_{2} O_{3} (s) + 3 C O (g) ⇌ 2 F e (s) + 3 C O_{2} (g); ∆ H^{\circ} = - 26.8 k J$

(II) $F e (s) + C O_{2} (g) ⇌ F e O (s) + C O (g); ∆ H^{\circ} = + 16.5 k J$

the $∆ H^{\circ}$ value, in kilojoules, for the reaction

$F e_{2} O_{3} (s) + C O (g) \to 2 F e O (s) + C O_{2} (g)$ is calculated to be:

0%

-43.3
0%

-10.3
0%

+6.2
0%

+10.3

82 litres of carbon dioxide are produced at a pressure of 1 atm by the action of acid on a metal carbonate. The work done at room temp by the gas (in calories) in pushing back the atmosphere is

0%

1000
0%

820
0%

1640
0%

2200

Among the following, the reaction for which $∆ H = ∆ E$ is-

0%

$P C l_{5} (g) \to P C l_{3} (g) + C l_{2} (g)$
0%

$H_{2} (g) + C l_{2} (g) \to 2 H C l (g)$
0%

$C_{2} H_{5} O H (l) + 3 O_{2} (g) \to 2 C O_{2} (g) + 3 H_{2} O (l)$
0%

$C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)$

For which change $∆ H \neq ∆ E$

0%

$H_{2} + I_{2} ⇌ 2 H I$
0%

$H C l + N a O H \to N a C l + H_{2} O$
0%

$C (s) + O_{2} (g) ⇌ C O_{2} (g)$
0%

$N_{2} + 3 H_{2} \to 2 N H_{3}$

Enthalpy change when 1.00 g water is frozen at $0^{\circ} C,$ is :

$(∆ H_{f u s} = 1.435 k c a l m o l^{- 1})$

0%

0.0797 kcal
0%

-0.0797 kcal
0%

1.435 kcal
0%

-1.435 kcal

One mole of hydrogen gas at $25^{\circ} C$ and 1 atm pressure is heated at constant pressure until its volume has doubled. Given that $C_{v}$ for hydrogen is 3.0 cal deg^-1 mol^-1, the $∆ H$ and $∆ E$ for this process are-

0%

$∆ H$ = 1490 cal and $∆ E$ = 894 cal
0%

$∆ H$ = $∆ E$ = 1490 cal
0%

$∆ H$ = 894 cal and $∆ E$ = 1490 cal
0%

$∆ H$ = $∆ E$ = 894 cal

The enthalpy change of the reaction

$A l_{2} C l_{6} (s) + 6 N a (s) \to 2 A l (s) + 6 N a C l (s)$ is -257 kcal.

Given the heat of formation of NaCl(s) is 98.0 kcal, the heat of formation of $A l_{2} C l_{6} (s)$ is-

0%

158 kcal
0%

-331 kcal
0%

-166 kcal
0%

316 kcal

Some of the thermodynamic parameters are state variables while some are process variables. Some grouping of the parameters are given. Choose the correct one-

Process variables : Internal energy, work done by the gas

Process variables : Temperature, volume

Process variables : Work done by the gas, heat absorbed by the gas

0%

State variables : Temperature, No. of moles
0%

State variables : Volume, Temperature
0%

State variables : work done by the gas, heat rejected by the gas.
0%

State variables : Internal energy, volume

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Practice Chemistry Quiz Questions and Answers

0:0:1

Practice Chemistry Quiz Questions and Answers

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