1
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
2
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.
3
MCQ (Single Correct Answer)

AIPMT 2008

Which of the following are not state functions ?
(I) q + w     (II) q
(III) w        (IV) H $$-$$ TS
A
(I), (II) and (III)
B
(II) and (III)
C
(I) and (IV)
D
(II), (III) and (IV)

Explanation

Enthalpy (H = q + W) and Gibbs energy, (G = H –TS) are state functions which depend only on the initial and final states of system.

While, heat (q) and work done (W) are the path function which depends on the path followed in bringing the change between two states of the system.
4
MCQ (Single Correct Answer)

AIPMT 2007

Given that bond energies of H $$-$$ H and Cl $$-$$ Cl are 430 kJ mol$$-$$1 and 240 kJ mol$$-$$1 respectively and $$\Delta $$Hf for HCl is $$-$$ 90 kJ mol$$-$$1, bond enthalpy of HCl is
A
380 kJ mol$$-$$1
B
425 kJ mol$$-$$1
C
245 kJ mol$$-$$1
D
290 kJ mol$$-$$1

Explanation

$${1 \over 2}$$H2 + $${1 \over 2}$$Cl2 $$ \to $$ HCl

$$\Delta $$Hf = –90 kJ mol–1

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

$$ \Rightarrow $$ -90 = [ $${1 \over 2}$$(430)H2 + $${1 \over 2}$$(240)Cl2] - (B.E)HCl

$$ \Rightarrow $$ -90 = [215 + 120] - (B.E)HCl

$$ \Rightarrow $$ (B.E)HCl = 425 kJ mol–1

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