1
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
2
MCQ (Single Correct Answer)

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.
3
MCQ (Single Correct Answer)

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.
4
MCQ (Single Correct Answer)

AIPMT 2008

Bond dissociation enthalpy of H2, Cl2 and HCl are 434, 242 and 431 kJ mol$$-$$1 respectively. Enthalpy of formation of HCl is
A
$$-$$ 93 kJ mol$$-$$1
B
245 kJ mol$$-$$1
C
93 kJ mol$$-$$1
D
$$-$$ 245 kJ mol$$-$$1

Explanation

$${1 \over 2}$$H2 + $${1 \over 2}$$Cl2 $$ \to $$ HCl, $$\Delta $$Hf = ?

$$\Delta $$Hreaction = [ $${1 \over 2}$$(B.E)H2 + $${1 \over 2}$$(B.E)Cl2] - (B.E)HCl

= [217 + 121] – 431

= 338 – 431 = – 93 kJ mol–1

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