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1

### AIPMT 2011 Prelims

MCQ (Single Correct Answer)
Enthalpy change for the reaction,
4H(g)  $$\to$$  2H2(g) is $$-$$869.6 kJ
The dissociation energy of H $$-$$ H bond is
A
434.8 kJ
B
$$-$$ 869.6 kJ
C
+ 434.8 kJ
D
+ 217.4 kJ

## Explanation

4H(g) → 2H2(g),    ∆H = – 869.6 kJ

Reverse the above equation

2H2(g) → 4H(g),    ∆H = + 869.6 kJ

Divide the above equation by 2,

H2(g) → 2H(g), $$\Delta H = {{869.6} \over 2}$$ kJ = 434.8 kJ
2

### AIPMT 2011 Prelims

MCQ (Single Correct Answer)
If the enthalpy change for the transition of liquid water to steam is 30 kJ mol$$-$$1 at 27oC, the entropy change for the process would be
A
10 J mol$$-$$1 K$$-$$1
B
1.0 J mol$$-$$1 K$$-$$1
C
0.1 J mol$$-$$1 K$$-$$1
D
100 J mol$$-$$1 K$$-$$1

## Explanation

H2O(l) $$\buildrel {300K} \over \longrightarrow$$ H2O(g)

∆H = 30 kJ mol–1

$$\Delta S = {{\Delta H} \over T}$$ = $${{30 \times {{10}^3}} \over {300}}$$

= 100 J mol–1 K–1
3

### AIPMT 2010 Mains

MCQ (Single Correct Answer)
The following two reactions are known

Fe2O3(s) + 3CO(g) $$\to$$ 2Fe(s) + 3CO2(g); $$\Delta$$H = $$-$$ 26.8 kJ

FeO(s) + CO(g) $$\to$$  Fe(s) + CO2(g); $$\Delta$$H = $$-$$ 16.5 kJ

The value of $$\Delta$$H for the following reaction

Fe2O3(s) + CO(g) $$\to$$  2FeO(s) + CO2(g) is
A
+ 10.3 kJ
B
$$-$$ 43.3 kJ
C
$$-$$ 10.3 kJ
D
+ 6.2 kJ

## Explanation

Given

Fe2O3(s) + 3CO(g) $$\to$$ 2Fe(s) + 3CO2(g); $$\Delta$$H = $$-$$ 26.8 kJ .....(1)

FeO(s) + CO(g) $$\to$$  Fe(s) + CO2(g); $$\Delta$$H = $$-$$ 16.5 kJ .....(2)

Fe2O3(s) + CO(g) $$\to$$  2FeO(s) + CO2(g), $$\Delta$$H = ? ....(3)

Equation (3) can be calculated as :

(1) - 2(2)

$$\therefore$$ $$\Delta$$H = –26.8 + 33.0 = +6.2 kJ
4

### AIPMT 2010 Mains

MCQ (Single Correct Answer)
For vaporization of water at 1 atmospheric pressure, the values of $$\Delta$$H and $$\Delta$$S are 40.63 kJ mol$$-$$1 and 108.8 J K$$-$$1 mol$$-$$1, respectively. The temperature when Gibb's energy change ($$\Delta$$G) for this transformation will be zero, is
A
273.4 K
B
393.4 K
C
373.4 K
D
293.4 K

## Explanation

We know, from Gibb's equation,

$$\Delta$$G = $$\Delta$$H – T$$\Delta$$S

When $$\Delta$$G = 0, $$\Delta$$H = T$$\Delta$$S

$$\therefore$$ $$T = {{\Delta H} \over {\Delta S}}$$ = $${{40.63 \times {{10}^3}} \over {108.8}}$$ = 373.4 K

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