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1

AIPMT 2004

MCQ (Single Correct Answer)
If the bond energies of H $$-$$ H, Br $$-$$ Br, and H $$-$$ Br are 433, 192 and 364 kJ mol$$-$$1 respectively, the $$\Delta $$Ho for the reaction

H2(g) + Br2(g) $$ \to $$ 2HBr(g) is
A
$$-$$ 261 kJ
B
+103 kJ
C
+261 kJ
D
$$-$$103 kJ

Explanation

H2(g) + Br2(g) $$ \to $$ 2HBr(g),   $$\Delta $$Hof = ?

$$\Delta $$Hof = $$\Sigma $$(B.E.)reactants – $$\Sigma $$(B.E.)products

= (B.E.)H–H + (B.E.)Br–Br –2(B.E)H-Br

= [433 + 192] – 2(364) kJ mol–1

= (625 – 728) kJ mol–1 = –103 kJ mol–1
2

AIPMT 2003

MCQ (Single Correct Answer)
What is the entropy change (in J K$$-$$1 mol$$-$$1) when one mole of ice is converted into water at 0oC? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol$$-$$1 at 0oC).
A
20.13
B
2.013
C
2.198
D
21.98

Explanation

H2O(s) → H2O(l)

$$\Delta $$H = 6.0 kJ mol–1

T = 0 + 273 K = 273 K; $$\Delta $$S = ?

$$\Delta $$S = $${{\Delta H} \over T}$$ = $${{6000} \over {273}}$$ = 21.98 JK–1 mol–1
3

AIPMT 2003

MCQ (Single Correct Answer)
The molar heat capacity of water at constant pressure, C, is 75 J K$$-$$1 mol$$-$$1. When 1.0 kJ of heat is supplied to 100 g of water which is free to expand, the increase in temperature of water is
A
1.2 K
B
2.4 K
C
4.8 K
D
6.6 K

Explanation

As we know, q = nC$$\Delta $$T

q = 1.0 kJ = 1000 J

C = 75 JK–1 mol–1

m = 100 g ⇒ Number of moles = $${{100} \over {18}}$$ g mol–1

1000 = $${{100} \over {18}}$$ $$ \times $$ 75 $$ \times $$ $$\Delta $$T

$$ \Rightarrow $$ $$\Delta $$T = $${{10 \times 18} \over {75}}$$ K

$$ \Rightarrow $$ $$\Delta $$T = 2.4 K
4

AIPMT 2003

MCQ (Single Correct Answer)
Formation of solution from two components can be considered as
(i) Pure solvent $$ \to $$ separated solvent molecules, $$\Delta $$H1
(ii) Pure solute $$ \to $$ separated solute molecules, $$\Delta $$H2
(iii) Separated solvent and solute molecules $$ \to $$ solution, $$\Delta $$H3
Solution so formed will be ideal if
A
$$\Delta $$Hsoln = $$\Delta $$H1 + $$\Delta $$H2 + $$\Delta $$H3
B
$$\Delta $$Hsoln = $$\Delta $$H1 + $$\Delta $$H2 $$-$$ $$\Delta $$H3
C
$$\Delta $$Hsoln = $$\Delta $$H1 $$-$$ $$\Delta $$H2 $$-$$ $$\Delta $$H3
D
$$\Delta $$Hsoln = $$\Delta $$H3 $$-$$ $$\Delta $$H1 $$-$$ $$\Delta $$H2

Explanation

For ideal solution,

$$\Delta $$Hsoln = $$\Delta $$H1 + $$\Delta $$H2 + $$\Delta $$H3

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