1
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

Which of the following is correct option for free expansion of an ideal gas under adiabatic condition ?
A
q = 0, $$\Delta $$T $$ \ne $$ 0, w = 0
B
q $$ \ne $$ 0,   $$\Delta $$T = 0,   w = 0
C
q = 0,   $$\Delta $$T = 0,   w = 0
D
q = 0,  $$\Delta $$T < 0, w $$ \ne $$ 0

Explanation

For adiabatic condition, change in heat does not take place thus, q = 0.

If q = 0 then change in temperature does not take place thus, $$\Delta $$T = 0.

Also for a free expansion of ideal gas work done, W = 0 as there is no external pressure on it.
2
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

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
3
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

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
4
MCQ (Single Correct Answer)

AIPMT 2010 Mains

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

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