1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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