1

### NEET 2016 Phase 1

Consider the following liquid-vapour equilibrium.
Liquid $\rightleftharpoons$ Vapour
Which of the following relations is correct ?
A
${{d\ln P} \over {d{T^2}}} = {{ - \Delta {H_v}} \over {{T^2}}}$
B
${{d\ln P} \over {dT}} = {{\Delta {H_v}} \over {R{T^2}}}$
C
${{d\ln G} \over {d{T^2}}} = {{\Delta {H_v}} \over {R{T^2}}}$
D
${{d\ln P} \over {dT}} = {{ - \Delta {H_v}} \over {RT}}$

## Explanation

This is Clausius-Clapeyron equation.

${{d\ln P} \over {dT}} = {{\Delta {H_v}} \over {R{T^2}}}$
2

### AIPMT 2015 Cancelled Paper

If the value of equilibrium constant for a particular reaction is 1.6 $\times$ 1012, then at equilibrium the system will contain
A
mostly products
B
similar amounts of reactants and products
C
all reactants
D
mostly reactants.

## Explanation

As, 1.6 × 1012 is very high value of K. Thus, the reaction proceeds almost to completion and mixture must contain mostly products.
3

### AIPMT 2015 Cancelled Paper

The Ksp of Ag2CrO4,  AgCl,  AgBr  and Agl  are respectively, 1.1 $\times$ 10$-$12, 1.8 $\times$ 10$-$10, 5.0 $\times$ 10$-$13, 8.3 $\times$ 10$-$17. Which one of the following salts will precipitate last if AgNO3 solution is added to the solution containing equal moles of NaCl, NaBr, Nal and Na2CrO4?
A
AgBr
B
Ag2CrO4
C
Agl
D
AgCl

## Explanation

From the Ksp values of the given salts calculate the solubility values. Salt having highest solubility will precipitate at last.
AgCrO4 2Ag+ + CrO42-
s 2s s

Ksp = (2s)2(s) = 1.1 × 10–12

$\Rightarrow$ s = 0.65 × 10–4

AgCl Ag+ + Cl-
s s s

Ksp = s × s

$\Rightarrow$ 1.8 × 10–10 = s 2

$\Rightarrow$ s = 1.34 × 10–5

AgBr Ag+ + Br-
s s s

Ksp = s × s

$\Rightarrow$ 5 × 10–13 = s2

$\Rightarrow$ s = 0.71 × 10–6

AgI Ag+ + I-
s s s

Ksp = s × s

$\Rightarrow$ 8.3 × 10–17 = s2

$\Rightarrow$ s = 0.9 × 10–8

$\therefore$ Solubility of Ag2CrO4 is maximum so, it will precipitate at last.
4

### AIPMT 2015 Cancelled Paper

Which of the following statements is correct for a reversible process in a state of equilibrium?
A
$\Delta$Go = $-$2.30 RT log K
B
$\Delta$Go = 2.30 RT log K
C
$\Delta$G = $-$2.30 RT log K
D
$\Delta$G = 2.30 RT log K

## Explanation

$\Delta$G = $\Delta$Go + 2.303 RT log Q

At equilibrium, when $\Delta$G = 0 and Q = K

then 0 = $\Delta$Go + 2.303 RT log K

$\Delta$Go = – 2.303 RT log K