1

### 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
2

### 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
3

### AIPMT 2009

The values of $\Delta$H and $\Delta$S for the reaction,

C(graphite) + CO2(g) $\to$  2CO(g)

are 170 kJ and 170 J K$-$1, respectively. This reaction will be spontaneous at
A
910 K
B
1110 K
C
510 K
D
710 K

## Explanation

We know, $\Delta$G = $\Delta$H – T$\Delta$S

For reaction to be spontaneous, $\Delta$G < 0

$\Rightarrow$ $\Delta$H – T$\Delta$S < 0

$\Rightarrow$ 170 $\times$ 103 - T(170) < 0

$\Rightarrow$ T > 1000 K

Among the given temperatures, only 1110 K is greater than 1000 K thus, at this temperate the reaction will be spontaneous.
4

### AIPMT 2008

For the gas phase reaction,

PCl5(g)  $\rightleftharpoons$ PCl3(g) + Cl2(g)

which of the following conditions are correct ?
A
$\Delta$H < 0 and $\Delta$S < 0
B
$\Delta$H > 0 and $\Delta$S < 0
C
$\Delta$H = 0 and $\Delta$S < 0
D
$\Delta$H > 0 and $\Delta$S > 0

## Explanation

$\Delta$H = $\Delta$E + $\Delta$ngRT

$\Delta$ng = (1 + 1) – (1) = 1

$\Rightarrow$ $\Delta$H = $\Delta$E + RT

Thus, $\Delta$H is a positive quantity i.e., ∆H > 0. Now one mole of gaseous reactant dissociate into two moles of gaseous products thus, entropy increases i.e., $\Delta$S > 0.