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1

### 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

### AIPMT 2003

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

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

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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