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1

AIPMT 2010 Mains

The following two reactions are known

Fe2O3(s) + 3CO(g) $$\to$$ 2Fe(s) + 3CO2(g); $$\Delta$$H = $$-$$ 26.8 kJ

FeO(s) + CO(g) $$\to$$  Fe(s) + CO2(g); $$\Delta$$H = $$-$$ 16.5 kJ

The value of $$\Delta$$H for the following reaction

Fe2O3(s) + CO(g) $$\to$$  2FeO(s) + CO2(g) is
A
+ 10.3 kJ
B
$$-$$ 43.3 kJ
C
$$-$$ 10.3 kJ
D
+ 6.2 kJ

Explanation

Given

Fe2O3(s) + 3CO(g) $$\to$$ 2Fe(s) + 3CO2(g); $$\Delta$$H = $$-$$ 26.8 kJ .....(1)

FeO(s) + CO(g) $$\to$$  Fe(s) + CO2(g); $$\Delta$$H = $$-$$ 16.5 kJ .....(2)

Fe2O3(s) + CO(g) $$\to$$  2FeO(s) + CO2(g), $$\Delta$$H = ? ....(3)

Equation (3) can be calculated as :

(1) - 2(2)

$$\therefore$$ $$\Delta$$H = –26.8 + 33.0 = +6.2 kJ
2

AIPMT 2010 Mains

For vaporization of water at 1 atmospheric pressure, the values of $$\Delta$$H and $$\Delta$$S are 40.63 kJ mol$$-$$1 and 108.8 J K$$-$$1 mol$$-$$1, respectively. The temperature when Gibb's energy change ($$\Delta$$G) for this transformation will be zero, is
A
273.4 K
B
393.4 K
C
373.4 K
D
293.4 K

Explanation

We know, from Gibb's equation,

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

When $$\Delta$$G = 0, $$\Delta$$H = T$$\Delta$$S

$$\therefore$$ $$T = {{\Delta H} \over {\Delta S}}$$ = $${{40.63 \times {{10}^3}} \over {108.8}}$$ = 373.4 K
3

AIPMT 2010 Mains

Three moles of an ideal gas expanded spontaneously into vacuum. The work done will be
A
infinite
B
3 Joules
C
9 Joules
D
zero

Explanation

Since the ideal gas expands spontaneously into vacuum,, Pext = 0.

$$\therefore$$ Work done is also zero.
4

AIPMT 2010 Mains

Match List I (Equations) with List II (Type of processes) and select the correct option.

List I List II
Equations Type of processes
A. Kp > Q (i) Non- spontaneous
B. $$\Delta$$Go < RT ln Q (ii) Equilibrium
C. Kp = Q (iii) Spontaneous and
endothermic
D. T > $${{\Delta H} \over {\Delta S}}$$ (iv) Spontaneous
A
A - (i), B - (ii), C - (iii), D - (iv)
B
A - (iii), B - (iv), C - (ii), D - (i)
C
A - (iv), B - (i), C - (ii), D - (iii)
D
A - (ii), B - (i), C - (iv), D - (iii)

Explanation

When Kp > Q, rate of forward reaction > rate of backward reaction.

$$\therefore$$ Reaction is spontaneous.

When $$\Delta$$Go < RT ln Q, $$\Delta$$Go is positive, reverse reaction is feasible, thus reaction is non spontaneous.

When Kp = Q, rate of forward reaction > rate of backward reaction.

$$\therefore$$ Reaction is in equilibrium.

When T$$\Delta$$S > $$\Delta$$H, $$\Delta$$G will be negative only when $$\Delta$$H = +ve.

$$\therefore$$ Reaction is spontaneous and endothermic.

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