1
MCQ (Single Correct Answer)

AIPMT 2004

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
MCQ (Single Correct Answer)

AIPMT 2004

Standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K are $$-$$ 382.64 kJ mol$$-$$1 and $$-$$ 145.6 kJ mol$$-$$1, respectively. Standard Gibb's energy change for the same reaction at 298 K is
A
$$-$$ 221.1 kJ mol$$-$$1
B
$$-$$339.3 kJ mol$$-$$1
C
$$-$$ 439.3 kJ mol$$-$$1
D
$$-$$ 523.2 kJ mol$$-$$1

Explanation

$$\Delta $$G = $$\Delta $$H – T$$\Delta $$S

$$\Delta $$G = –382.64 × 103 J mol–1 – (298K) (–145.6 JK–1 mol–1)

= –382640 + 43388.8

= – 339251.2 J mol–1 = – 339.3 kJ mol–1
3
MCQ (Single Correct Answer)

AIPMT 2004

The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm is (1 L atm = 101.32 J)
A
$$-$$ 6 J
B
$$-$$ 608 J
C
+ 304 J
D
$$-$$ 304 J

Explanation

Work done during the expansion, W = – pdV

W = –3 atm (6 dm3 – 4 dm3)

= – 3 atm ( 2 dm3 ) (1 dm3 = 1 L)

= – 3 atm × 2 L

= – 6 L atm

As, 1 L atm = 101.32 J

$$ \therefore $$ W = – 6 × 101.32 J = – 607.92 J ≈ – 608 J
4
MCQ (Single Correct Answer)

AIPMT 2003

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

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