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1

NEET 2013

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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.

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