1
MCQ (Single Correct Answer)

AIPMT 2014

For a given exothermic reaction, Kp and K'p are the equilibrium constants at temperatures T1 and T2, respectively. Assuming that heat of reaction is constant in temperature range between T1 and T2, it is readily observed that
A
Kp > K'p
B
Kp < K'p
C
Kp = K'p
D
Kp = $${1 \over {k{'_p}}}$$

Explanation

log
K'p
Kp
= $$ - {{\Delta H} \over {2.303R}}\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]$$

For exothermic reaction, $$\Delta $$H = -ve means the temperature T2 is higher than T1.

$$ \therefore $$ $$\left[ {{1 \over {{T_2}}} - {1 \over {{T_1}}}} \right]$$ is negative.

So log K'p - log Kp = -ve

$$ \Rightarrow $$ log Kp > log K'p

$$ \Rightarrow $$ Kp > K'p
2
MCQ (Single Correct Answer)

AIPMT 2014

Using the Gibb's energy change, $$\Delta $$Go = +63.3 kJ, for the following reaction,

Ag2CO3(s) $$\rightleftharpoons$$ 2 Ag+(aq) + CO32$$-$$ (aq)
the Ksp of Ag2CO3(s) in water at 25oC is
(R = 8.314 J K$$-$$1 mol$$-$$1)
A
3.2 $$ \times $$ 10$$-$$26
B
8.0 $$ \times $$ 10$$-$$12
C
2.9 $$ \times $$ 10$$-$$3
D
7.9 $$ \times $$ 10$$-$$2

Explanation

We know, $$\Delta $$Go = – 2.303 RT log Ksp

$$ \therefore $$ 63300 = – 2.303 × 8.314 × 298 log Ksp

$$ \Rightarrow $$ log Ksp = -11.09

$$ \Rightarrow $$ Ksp = 10-11.09

= 8.0 × 10–12
3
MCQ (Single Correct Answer)

AIPMT 2014

Which of the following salts will give highest pH in water?
A
KCl
B
NaCl
C
Na2CO3
D
CuSO4

Explanation

Na2CO3 is salt of strong base, NaOH and weak acid, H2CO3 hence the pH value of the solution will be high.
4
MCQ (Single Correct Answer)

NEET 2013 (Karnataka)

The dissociation constant of weak acid is 1 $$ \times $$ 10$$-$$4. In order to prepare a buffer solution with a pH = 5, the [Salt]/[Acid] ratio should be
A
4 : 5
B
10 : 1
C
5 : 4
D
1 : 10

Explanation

Given, Ka = 1 × 10–4

pKa = – log (1× 10–4) = 4

pH = pKa + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ 5 = 4 + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 1

$$ \Rightarrow $$ $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$ = 10 = 10 : 1

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