1
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

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
2
MCQ (Single Correct Answer)

NEET 2013

A reaction having equal energies of activation for forward and reverse reactions has
A
$$\Delta $$H = 0
B
$$\Delta $$H = $$\Delta $$G = $$\Delta $$S = 0
C
$$\Delta $$S = 0
D
$$\Delta $$G = 0

Explanation

For a general reaction,

ΔH = Activation energy of forward reaction – Activation energy of backward reaction.

As, both the energies of activation have same value thus, ΔH = 0.

$$\Delta $$G is not equal to zero because if it is so the reaction must be in equilibrium which is not in this case
3
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0oC is
A
10.52 cal/(mol K)
B
21.04 cal/(mol K)
C
5.260 cal/(mol K)
D
0.526 cal/(mol K)

Explanation

H2O($$l$$) → H2O(s)

∆H = 1.435 Kcal/mol

T = 0 + 273K = 273K

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

$$ \Rightarrow $$ $$\Delta S = {{1.435} \over {273}}$$ = 5.26 $$ \times $$ 10-3 kcal/mol K

$$ \Rightarrow $$ $$\Delta S$$ = 5.260 cal/mol K
4
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

Standard enthalpy of vaporisation $$\Delta $$vapHo for water at 100oC is 40.66 kJ mol$$-$$1. The internal energy of vaporisation of water at 100oC (in kJ mol$$-$$1) is
A
+37.56
B
$$-$$43.76
C
+ 43.76
D
+ 40.66

Explanation

H2O(l) $$\buildrel {100^\circ \,C} \over \longrightarrow $$ H2O(g)

∆Ho = 40.66kJ mol–1

∆Ho = ∆uo + $$\Delta $$ng RT

$$\Delta $$ng = 1, R = 8.314 × 10–3 kJ mol–1 k–1

T = 100 + 273 = 373 K

$$ \Rightarrow $$ 40.66 = ∆uo + (1) (8.314 × 10–3) × 373

$$ \Rightarrow $$ ∆uo = 37.56 kJ mol–1

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