1
MCQ (Single Correct Answer)

AIPMT 2001

Change in enthalpy for reaction,

2H2O2(l) $$ \to $$ 2H2O(l) + O2(g)

if heat of formation of H2O2(l) and H2O(l) are $$-$$188 and - 286 kJ/mol respectively, is
A
$$-$$ 196 kJ/mol
B
+ 196 kJ/mol
C
+948 kJ/mol
D
$$-$$ 948 kJ/mol

Explanation

2H2O2(l) $$ \to $$ 2H2O(l) + O2(g), $$\Delta $$Hr = ?

H2(g) + O2(g) → H2O2(l), $$\Delta $$H = – 188 kJ mol ...(1)

H2(g) + O2(g) → H2O(l), $$\Delta $$H = –286 kJ/mol ....(2)

(1) - (2)

H2O2(l) → H2O(l) + $${1 \over 2}$$O2(g)

$$\Delta $$H = –286 – (– 188) = –98 kJ mol–1

Multiplying this equation by 2 we get the required equation

2H2O2(l) → 2H2O(l) + O2(g)

$$\Delta $$H = 2 (– 98) kJ/mol = – 196 kJ/mol
2
MCQ (Single Correct Answer)

AIPMT 2000

For the reaction,
C2H5OH(l) + 3O2(g) $$ \to $$ 2CO2(g) + 3H2O(l)
which one is true
A
$$\Delta $$H = $$\Delta $$E $$-$$RT
B
$$\Delta $$H = $$\Delta $$E + RT
C
$$\Delta $$H = $$\Delta $$E + 2RT
D
$$\Delta $$H = $$\Delta $$E $$-$$ 2RT

Explanation

As we know, $$\Delta $$H = $$\Delta $$E + $$\Delta $$ng RT

Now, $$\Delta $$ng = Number of gaseous moles of products – number of gaseous moles of reactions

= 2 – 3 = – 1

$$ \Rightarrow $$ $$\Delta $$H = $$\Delta $$E + (–1) RT

$$\Delta $$H = $$\Delta $$E – RT
3
MCQ (Single Correct Answer)

AIPMT 2000

The entropy change in the fusion of one mole of a solid melting at 27oC (latent heat of fusion is 2930 J mol–1) is :
A
9.77 J/mol-K
B
10.77 J/mol-K
C
9.07 J/mol-K
D
0.977 J/mol-K

Explanation

$$\Delta S = {{\Delta H} \over T}$$

= $${{2930} \over {300}}$$

= 9.77 J/mol-K
4
MCQ (Single Correct Answer)

AIPMT 2000

2Zn + O2 $$ \to $$ 2ZnO;  $$\Delta $$Go = $$-$$ 616 J
2Zn + S2 $$ \to $$ 2ZnS;  $$\Delta $$Go = $$-$$ 293 J
S2 + 2O2 $$ \to $$ 2SO2;   $$\Delta $$Go = $$-$$408 J
$$\Delta $$Go for the following reaction
2ZnS + 3O2 $$ \to $$  2ZnO + 2SO2 is
A
$$-$$ 731 J
B
$$-$$ 1317 J
C
$$-$$ 501 J
D
+ 731 J

Explanation

2Zn + O2 $$ \to $$ 2ZnO;  $$\Delta $$Go = $$-$$ 616 J ....(1)
2ZnS $$ \to $$ 2Zn + O2;  $$\Delta $$Go = $$+$$ 293 J.....(2)
S2 + 2O2 $$ \to $$ 2SO2;   $$\Delta $$Go = $$-$$408 J .....(3)

$$\Delta $$Go for the reaction can be obtained by adding (1), (2) and (3)

$$ \therefore $$ $$\Delta $$Go = 293 - 616 - 408 = -731 J

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