1
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
2
MCQ (Single Correct Answer)

AIPMT 2007

Consider the following reactions :
(i) H+(aq) + OH$$-$$(aq) = H2O(l),  $$\Delta $$H = $$-$$ X1 kJ mol$$-$$1
(ii) H2(g) + 1/2O2(g) = H2O(l),   $$\Delta $$H = $$-$$ X2 kJ mol$$-$$1
(iii) CO2(g) + H2(g) = CO(g) + H2O(l),  $$\Delta $$H = $$-$$ X3 kJ mol$$-$$1
(iv) C2H2(g) + 5/2O2(g) = 2CO2(g) + H2O(l),    $$\Delta $$H = +X4 kJ mol$$-$$1

Enthalpy of formation of H2O(l) is
A
+X3 kJ mol$$-$$1
B
$$-$$X4 kJ mol$$-$$1
C
+X1 kJ mol$$-$$1
D
$$-$$X2 kJ mol$$-$$1

Explanation

Chemical equation for the formation of H2O(l) is

H2(g) + 1/2O2(g) = H2O(l)

Because enthalpy of formation of a compound is the heat absorbed or released when one mole of this substance is formed from its constituent elements.

Thus, enthalpy of formation of H2O is –X2 kJ mol–1 where negative sign shows that the reaction is exothermic.

Equation (i) represents nutralisation reaction.

Equation (iii) represents hydrogenation reaction.

Equation (iv) represents combustion reaction.
3
MCQ (Single Correct Answer)

AIPMT 2006

The enthalpy of hydrogenation of cyclohexene is is $$-$$ 119.5 kJ mol$$-$$1. If resonance energy of benzene is $$-$$ 150.4 kJ mol$$-$$1, its enthalpy of hydrogenation would be
A
$$-$$ 358.5 kJ mol$$-$$1
B
$$-$$ 508.9 kJ mol$$-$$1
C
$$-$$ 208.1 kJ mol$$-$$1
D
$$-$$ 269.9 kJ mol$$-$$1

Explanation



The resonance energy provides extra stability to the benzene molecule so it has to over come for hydrogenation to take place.

So $$\Delta $$H = – 358.5 – (–150.4) = –208.1 kJ
4
MCQ (Single Correct Answer)

AIPMT 2006

Assume each reaction is carried out in an open container. For which reaction will $$\Delta $$H = $$\Delta $$E ?
A
2CO(g) + O2(g) $$ \to $$  2CO2(g)
B
H2(g) + Br2(g) $$ \to $$ 2HBr(g)
C
C(s) + 2H2O(g) $$ \to $$ 2H2(g) + CO2(g)
D
PCl5(g) $$ \to $$ PCl3(g) + Cl2(g)

Explanation

We know that

$$\Delta $$H = $$\Delta $$E + $$\Delta $$ngRT

In the reaction, H2(g) + Br2(g) $$ \to $$ 2HBr(g)

$$\Delta $$ng = 2 - (1 + 1) = 0

So, $$\Delta $$H = $$\Delta $$E for this reaction.

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