1

### 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.
2

### AIPMT 2010 Prelims

Standard entropies of X2, Y2 and XY3 are 60, 40 and 50 J K$-$1 mol$-$1 respectively. For the reaction

1/2X2 + 3/2Y2 $\rightleftharpoons$ XY3, $\Delta$H = $-$ 30 kJ,

to be at equilibrium, the temperature should be
A
750 K
B
1000 K
C
1250 K
D
500 K

## Explanation

Given reaction is :

${1 \over 2}$X2 + ${3 \over 2}$Y2 ⇌ XY3

We know,

$\Delta$So = $\sum {S_{products}^o} - \sum {S_{reac\tan ts}^o}$

= 50 - (30 + 60) = -40 J K-1 mol-1

At equilibrium $\Delta$Go = 0

$\Delta$Ho = T$\Delta$So

$\therefore$ $T = {{\Delta {H^o}} \over {\Delta {S^o}}}$ = ${{ - 30 \times {{10}^3}} \over { - 40}}$ = 750 K
3

### AIPMT 2010 Prelims

For an endothermic reaction, energy of activation is Ea and enthalpy of reaction is $\Delta$H (both of these in kJ/mol). Minimum value of Ea will be
A
less than $\Delta$H
B
equal to $\Delta$H
C
more than $\Delta$H
D
equal to zero

## Explanation

Here,

Ea = activation energy of forward reaction

E’a = activation energy of backward reaction

$\Delta$H = enthalpy of the reaction From the given diagram it is clear that

Ea = E’a + $\Delta$H

$\therefore$ Ea > $\Delta$H
4

### AIPMT 2009

From the following bond energies :
H $-$ H bond energy   : 431.37 kJ mol$-$1
C $=$ C bond energy   : 606.10 kJ mol$-$1
C $-$ C bond energy   : 336.49 kJ mol$-$1
C $-$ H bond energy   : 410.50 kJ mol$-$1
Enthalpy for the reaction, will be
A
$-$ 243.6 kJ mol$-$1
B
$-$ 120.0 kJ mol$-$1
C
553.0 kJ mol$-$1
D
1523.6 kJ mol$-$1

## Explanation

$\Delta$Hreaction = Σ(Bond enthalpy)reactants

– Σ(Bond enthalpy)products

= [B.E(C-C) + B.E(H-H) + 4$\times$B.E(C-H)]

- [B.E(C-C) + 6$\times$B.E(C-H)]

= [606.10 + 4(410.50) + 431.37]

– [336.49 + 6(410.50)]

= 2679.47 – 2799.49

= – 120.02 kJ mol–1