1
MCQ (Single Correct Answer)

NEET 2017

For a given reaction, $$\Delta $$H = 35.5 kJ mol$$-$$1 and $$\Delta $$S = 83.6 J K$$-$$1 mol$$-$$1. The reaction is spontaneous at (Assume that $$\Delta $$H and $$\Delta $$S do not vary with temperature.)
A
T > 425 K
B
all temperatures
C
T > 298 K
D
T < 425 K

Explanation

For a spontaneous reaction,

$$\Delta $$G < 0 i.e. $$\Delta $$H - T$$\Delta $$S < 0

$$T > {{\Delta H} \over {\Delta S}}$$

$$T > \left( {{{35.5 \times 1000} \over {83.6}} = 424.6 \approx 425K} \right)$$

$$ \therefore T > 425\,K$$
2
MCQ (Single Correct Answer)

NEET 2016 Phase 2

For a sample of perfect gas when its pressure is changed isothermally from pi to pf, the entropy change is given by
A
$$\Delta S = nR\,\ln \left( {{{{p_f}} \over {{p_i}}}} \right)$$
B
$$\Delta S = nR\,\ln \left( {{{{p_i}} \over {{p_f}}}} \right)$$
C
$$\Delta S = nRT\,\ln \left( {{{{p_f}} \over {{p_i}}}} \right)$$
D
$$\Delta S = RT\,\ln \left( {{{{p_i}} \over {{p_f}}}} \right)$$

Explanation

For an ideal gas undergoing reversible expansion, when temperature changes from Ti to Tf and pressure changes from Pi to Pf.

$$\Delta $$S = nCp ln $${{{T_f}} \over {{T_i}}}$$ + nR ln $${{{P_i}} \over {{P_f}}}$$

For an isothermal process, Ti = Tf so, ln 1 = 0

$$ \therefore $$ $$\Delta $$S = nR ln $${{{P_i}} \over {{P_f}}}$$
3
MCQ (Single Correct Answer)

AIPMT 2015

The heat of combination of carbon to CO2 is $$-$$393.5 kJ/mol. The heat released upon formation of 35.2 g of CO2 from carbon and oxygen gas is
A
+ 315 kJ
B
$$-$$630 kJ
C
$$-$$ 3.15 kJ
D
$$-$$ 315 kJ

Explanation

Given
$${C_{(s)}} + {O_{2(g)}} \to C{O_{(g)}},\Delta H = - 393.5\,kJ/mol$$

Then amount of heat released on formation of 44 g CO2 = 393.5 kJ

$$ \therefore $$ Amount of heat released on formation of

35.2 g CO2 = $${{393.5} \over {44}} \times 35.2 = 314.8 \approx 315\,kJ$$
4
MCQ (Single Correct Answer)

AIPMT 2014

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

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

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

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