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1

AIPMT 2014

MCQ (Single Correct Answer)
For the reaction, $${X_2}{O_{4\left( l \right)}}\,\, \to \,\,2X{O_{2(g)}}$$
$$\Delta $$U = 2.1 kcal, $$\Delta $$S = 20 cal K$$-$$1 at 300 K
A
2.7 kcal
B
$$-$$ 2.7 kcal
C
9.3 kcal
D
$$-$$ 9.3 kcal

Explanation

$$\Delta H = \Delta U + \Delta {n_g}RT$$

Given $$\Delta U = 2.1\,kcal,\,\Delta {n_g} = 2$$

$$R = 2 \times {10^{ - 3}}kcal,T = 300K$$

$$ \therefore $$ $$\Delta H = 2.1 + 2 \times 2 \times {10^{ - 3}} \times 300 = 3.3\,kcal$$

Again, $$\Delta G = \Delta H + T\Delta S$$

Given $$\Delta $$S = 20 $$ \times $$ 10-3 kcal K-1

On putting the values of $$\Delta $$H and $$\Delta $$S in the equation, we get

$$\Delta G = 3.3 - 300 \times 20 \times {10^{ - 3}}$$

$$ \Rightarrow 3.3 - 6 \times {10^3} \times {10^{ - 3}} = - 2.7\,kcal$$
2

AIPMT 2014

MCQ (Single Correct Answer)
Which of the following statements is correct for the spontaneous adsorption of a gas?
A
$$\Delta $$S is negative and, therefore $$\Delta $$H should be highly positive.
B
$$\Delta $$S is negative and therefore, $$\Delta $$H should be highly negative.
C
$$\Delta $$S is positive and therefore, $$\Delta $$H should be negative .
D
$$\Delta $$S is positive and therefore, $$\Delta $$H should also be highly positive.

Explanation

Using Gibb's-Helmholtz equation,

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

During adsorption of a gas, entropy decreases i.e, $$\Delta $$S < 0

For spontaneous adsorption, $$\Delta $$G should be negative, which is possible when $$\Delta $$H is highly negative.
3

NEET 2013 (Karnataka)

MCQ (Single Correct Answer)
Three thermochemical equations are given below
(i)  C(graphite) + O2(g) $$ \to $$ CO2(g); $$\Delta $$rHo = x kJ mol$$-$$1
(ii)  C(graphite) + $${1 \over 2}$$O2(g) $$ \to $$ CO(g); $$\Delta $$rHo = y kJ mol$$-$$1
(iii)  CO(g) + $${1 \over 2}$$O2(g) $$ \to $$ CO2(g); $$\Delta $$rHo = z kJ mol$$-$$1
Based on the above equations, find out which of the relationship given below is correct.
A
z = x + y
B
x = y + z
C
y = 2z $$-$$ x
D
x = y $$-$$ z

Explanation

According to Hess's law, equation (i) is equal to equations (ii) + (iii)
4

NEET 2013 (Karnataka)

MCQ (Single Correct Answer)
When 5 litres of a gas mixture of methane and propane is perfectly combusted at 0oC and 1 atmosphere, 16 litres of oxygen at the same temperature and pressure is consumed, The amount of heat released from this combustion in kJ ($$\Delta $$Hcomb. (CH4) = 890 kJ mol$$-$$1, $$\Delta $$Hcomb. (C3H8) = 2220 kJ mol$$-$$1) is
A
38
B
317
C
477
D
32

Explanation

CH4 + 2O2 $$ \to $$ CO2 + 2H2O

C3H8 + 5O2 $$ \to $$ 3CO2 + 4H2O

CH4 + C3H8 = $${5 \over {22.4}} = 0.22$$ moles.

$${O_2} = {{16} \over {22.4}} = 0.71$$ moles

$$2x + (0.22 \times x)5 = 0.71$$

x = 0.13

Heat liberated = 0.13 $$ \times $$ 890 + 0.09 $$ \times $$ 2220 = 316 kJ

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