1

### NEET 2013

A reaction having equal energies of activation for forward and reverse reactions has
A
$\Delta$H = 0
B
$\Delta$H = $\Delta$G = $\Delta$S = 0
C
$\Delta$S = 0
D
$\Delta$G = 0

## Explanation

For a general reaction,

ΔH = Activation energy of forward reaction – Activation energy of backward reaction.

As, both the energies of activation have same value thus, ΔH = 0.

$\Delta$G is not equal to zero because if it is so the reaction must be in equilibrium which is not in this case
2

### AIPMT 2012 Prelims

The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0oC is
A
10.52 cal/(mol K)
B
21.04 cal/(mol K)
C
5.260 cal/(mol K)
D
0.526 cal/(mol K)

## Explanation

H2O($l$) → H2O(s)

∆H = 1.435 Kcal/mol

T = 0 + 273K = 273K

$\Delta S = {{\Delta H} \over T}$

$\Rightarrow$ $\Delta S = {{1.435} \over {273}}$ = 5.26 $\times$ 10-3 kcal/mol K

$\Rightarrow$ $\Delta S$ = 5.260 cal/mol K
3

### AIPMT 2012 Prelims

Standard enthalpy of vaporisation $\Delta$vapHo for water at 100oC is 40.66 kJ mol$-$1. The internal energy of vaporisation of water at 100oC (in kJ mol$-$1) is
A
+37.56
B
$-$43.76
C
+ 43.76
D
+ 40.66

## Explanation

H2O(l) $\buildrel {100^\circ \,C} \over \longrightarrow$ H2O(g)

∆Ho = 40.66kJ mol–1

∆Ho = ∆uo + $\Delta$ng RT

$\Delta$ng = 1, R = 8.314 × 10–3 kJ mol–1 k–1

T = 100 + 273 = 373 K

$\Rightarrow$ 40.66 = ∆uo + (1) (8.314 × 10–3) × 373

$\Rightarrow$ ∆uo = 37.56 kJ mol–1
4

### AIPMT 2012 Prelims

In which of the following reactions, standard reaction entropy change ($\Delta$So) is positive and standard Gibb's energy change ($\Delta$Go) decreases sharply with increasing temperature ?
A
C(graphite) + ${1 \over 2}$O2(g) $\to$ CO(g)
B
CO(g) + ${1 \over 2}$O2(g) $\to$ CO2(g)
C
Mg(s) + ${1 \over 2}$O2(g) $\to$ MgO(g)
D
${1 \over 2}$C(graphite) + ${1 \over 2}$O2(g) $\to$ ${1 \over 2}$CO2(g)

## Explanation

$\Delta$So positive means entropy of products is more than that of reactants. Among the given reactions only in option (a) the $\Delta$So is positive cause products has 1 mole of gaseous products CO while reaction have half mole of gaseous O2 and Carbon (C) in solid state. Thus, reactions have less randomness than products. Also, the reaction is a combustion reaction which has a negative value of $\Delta$H. According to Gibbs energy change ($\Delta$So).

$\Delta$Go = $\Delta$Ho - T$\Delta$So

$\Rightarrow$ $\Delta$Go = -ve - T(+ve)

Thus, on increasing temperature value of $\Delta$Go decreases sharply.