1

### 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
2

### AIPMT 2010 Mains

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
3

### AIPMT 2010 Mains

Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be
A
infinite
B
3 Joules
C
9 Joules
D
zero

## Explanation

Since the ideal gas expands spontaneously into vacuum,, Pext = 0.

$\therefore$ Work done is also zero.
4

### AIPMT 2010 Mains

Match List I (Equations) with List II (Type of processes) and select the correct option.

List I List II
Equations Type of processes
A. Kp > Q (i) Non- spontaneous
B. $\Delta$Go < RT ln Q (ii) Equilibrium
C. Kp = Q (iii) Spontaneous and
endothermic
D. T > ${{\Delta H} \over {\Delta S}}$ (iv) Spontaneous
A
A - (i), B - (ii), C - (iii), D - (iv)
B
A - (iii), B - (iv), C - (ii), D - (i)
C
A - (iv), B - (i), C - (ii), D - (iii)
D
A - (ii), B - (i), C - (iv), D - (iii)

## Explanation

When Kp > Q, rate of forward reaction > rate of backward reaction.

$\therefore$ Reaction is spontaneous.

When $\Delta$Go < RT ln Q, $\Delta$Go is positive, reverse reaction is feasible, thus reaction is non spontaneous.

When Kp = Q, rate of forward reaction > rate of backward reaction.

$\therefore$ Reaction is in equilibrium.

When T$\Delta$S > $\Delta$H, $\Delta$G will be negative only when $\Delta$H = +ve.

$\therefore$ Reaction is spontaneous and endothermic.