1
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

In which of the following reactions, standard reaction entropy change ($$\Delta $$So) is positive and standard Gibb's energy change ($$\Delta $$Go) decreases sharply with increasing temperature ?
A
C(graphite) + $${1 \over 2}$$O2(g) $$ \to $$ CO(g)
B
CO(g) + $${1 \over 2}$$O2(g) $$ \to $$ CO2(g)
C
Mg(s) + $${1 \over 2}$$O2(g) $$ \to $$ MgO(g)
D
$${1 \over 2}$$C(graphite) + $${1 \over 2}$$O2(g) $$ \to $$ $${1 \over 2}$$CO2(g)

Explanation

$$\Delta $$So positive means entropy of products is more than that of reactants. Among the given reactions only in option (a) the $$\Delta $$So is positive cause products has 1 mole of gaseous products CO while reaction have half mole of gaseous O2 and Carbon (C) in solid state. Thus, reactions have less randomness than products. Also, the reaction is a combustion reaction which has a negative value of $$\Delta $$H. According to Gibbs energy change ($$\Delta $$So).

$$\Delta $$Go = $$\Delta $$Ho - T$$\Delta $$So

$$ \Rightarrow $$ $$\Delta $$Go = -ve - T(+ve)

Thus, on increasing temperature value of $$\Delta $$Go decreases sharply.
2
MCQ (Single Correct Answer)

AIPMT 2011 Mains

Consider the following processes :

H (kJ/mol)
1/2A $$ \to $$ B +150
3B  $$ \to $$ 2C + D -125
E + A  $$ \to $$ 2D +350

For B + D $$ \to $$ E + 2C, $$\Delta $$H will be
A
525 kJ/mol
B
$$-$$175 kJ/mol
C
$$-$$ 325 kJ/mol
D
325 kJ/mol

Explanation

H (kJ/mol)
1/2A $$ \to $$ B +150
3B  $$ \to $$ 2C + D -125
E + A  $$ \to $$ 2D +350
___________________________________
   B + D $$ \to $$ E + 2C;

$$\Delta $$H = (300 - 125 - 350) = - 175 kJ/mol
3
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

Which of the following is correct option for free expansion of an ideal gas under adiabatic condition ?
A
q = 0, $$\Delta $$T $$ \ne $$ 0, w = 0
B
q $$ \ne $$ 0,   $$\Delta $$T = 0,   w = 0
C
q = 0,   $$\Delta $$T = 0,   w = 0
D
q = 0,  $$\Delta $$T < 0, w $$ \ne $$ 0

Explanation

For adiabatic condition, change in heat does not take place thus, q = 0.

If q = 0 then change in temperature does not take place thus, $$\Delta $$T = 0.

Also for a free expansion of ideal gas work done, W = 0 as there is no external pressure on it.
4
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

Enthalpy change for the reaction,
4H(g)  $$ \to $$  2H2(g) is $$-$$869.6 kJ
The dissociation energy of H $$-$$ H bond is
A
434.8 kJ
B
$$-$$ 869.6 kJ
C
+ 434.8 kJ
D
+ 217.4 kJ

Explanation

4H(g) → 2H2(g),    ∆H = – 869.6 kJ

Reverse the above equation

2H2(g) → 4H(g),    ∆H = + 869.6 kJ

Divide the above equation by 2,

H2(g) → 2H(g), $$\Delta H = {{869.6} \over 2}$$ kJ = 434.8 kJ

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