1

### AIPMT 2014

For the reaction, ${X_2}{O_{4\left( l \right)}}\,\, \to \,\,2X{O_{2(g)}}$
$\Delta$U = 2.1 kcal, $\Delta$S = 20 cal K$-$1 at 300 K
A
2.7 kcal
B
$-$ 2.7 kcal
C
9.3 kcal
D
$-$ 9.3 kcal

## Explanation

$\Delta H = \Delta U + \Delta {n_g}RT$

Given $\Delta U = 2.1\,kcal,\,\Delta {n_g} = 2$

$R = 2 \times {10^{ - 3}}kcal,T = 300K$

$\therefore$ $\Delta H = 2.1 + 2 \times 2 \times {10^{ - 3}} \times 300 = 3.3\,kcal$

Again, $\Delta G = \Delta H + T\Delta S$

Given $\Delta$S = 20 $\times$ 10-3 kcal K-1

On putting the values of $\Delta$H and $\Delta$S in the equation, we get

$\Delta G = 3.3 - 300 \times 20 \times {10^{ - 3}}$

$\Rightarrow 3.3 - 6 \times {10^3} \times {10^{ - 3}} = - 2.7\,kcal$
2

### AIPMT 2014

Which of the following statements is correct for the spontaneous adsorption of a gas?
A
$\Delta$S is negative and, therefore $\Delta$H should be highly positive.
B
$\Delta$S is negative and therefore, $\Delta$H should be highly negative.
C
$\Delta$S is positive and therefore, $\Delta$H should be negative .
D
$\Delta$S is positive and therefore, $\Delta$H should also be highly positive.

## Explanation

Using Gibb's-Helmholtz equation,

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

During adsorption of a gas, entropy decreases i.e, $\Delta$S < 0

For spontaneous adsorption, $\Delta$G should be negative, which is possible when $\Delta$H is highly negative.
3

### NEET 2013 (Karnataka)

Three thermochemical equations are given below
(i)  C(graphite) + O2(g) $\to$ CO2(g); $\Delta$rHo = x kJ mol$-$1
(ii)  C(graphite) + ${1 \over 2}$O2(g) $\to$ CO(g); $\Delta$rHo = y kJ mol$-$1
(iii)  CO(g) + ${1 \over 2}$O2(g) $\to$ CO2(g); $\Delta$rHo = z kJ mol$-$1
Based on the above equations, find out which of the relationship given below is correct.
A
z = x + y
B
x = y + z
C
y = 2z $-$ x
D
x = y $-$ z

## Explanation

According to Hess's law, equation (i) is equal to equations (ii) + (iii)
4

### NEET 2013 (Karnataka)

When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0oC and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed, The amount of heat released from this combustion in kJ ($\Delta$Hcomb. (CH4) = 890 kJ mol$-$1, $\Delta$Hcomb. (C3H8) = 2220 kJ mol$-$1) is
A
38
B
317
C
477
D
32

## Explanation

CH4 + 2O2 $\to$ CO2 + 2H2O

C3H8 + 5O2 $\to$ 3CO2 + 4H2O

CH4 + C3H8 = ${5 \over {22.4}} = 0.22$ moles.

${O_2} = {{16} \over {22.4}} = 0.71$ moles

$2x + (0.22 \times x)5 = 0.71$

x = 0.13

Heat liberated = 0.13 $\times$ 890 + 0.09 $\times$ 2220 = 316 kJ