1

### JEE Main 2016 (Online) 9th April Morning Slot

The plot shows the variation of −$ln$ Kp versus temperature for the two reactions.

M(s) + ${1 \over 2}$ O2(g) $\to$ MO(s) and

C(s) + ${1 \over 2}$ O2(g) $\to$ CO(s) Identify the correct statement :
A
At T > 1200 K, carbon will reduce MO(s) to M(s).
B
At T < 1200 K, the reaction

MO(s) + C(s) $\to$ M(s) + CO(g) is spontaneous.
C
At T < 1200 K, oxidation of carbon is unfavourable
D
Oxidation of carbon is favourable at all temperatures.
2

### JEE Main 2016 (Online) 10th April Morning Slot

A solid XY kept in an evacuated sealed container undergoes decomposition to form a mixture of gases X and Y at temperature T. The equilibrium pressure is 10 bar in this vessel. Kp for this reaction is :
A
5
B
10
C
25
D
100
3

### JEE Main 2017 (Offline)

pKa of a weak acid (HA) and pKb of a weak base (BOH) are 3.2 and 3.4, respectively. The pH of their salt (AB) solution is:
A
6.9
B
7.0
C
1.0
D
7.2

## Explanation

The salt (AB) given is a salt is of weak acid and weak base. Hence its pH can be calculated by the formula

$\therefore$ pH = 7 + ${1 \over 2}p{K_a} - {1 \over 2}p{K_b}$

= 7 + ${1 \over 2}\left( {3.2} \right) - {1 \over 2}\left( {3.4} \right)$

= 6.9
4

### JEE Main 2017 (Online) 8th April Morning Slot

Additin of sodium hydroxide solution to a weak acid (HA)results in a buffer of pH 6. If ionition constant of HA is 10$-$5, the ratio of salt to acid concentration in the buffer solution will be :
A
4 : 5
B
1 : 10
C
10 : 1
D
5 : 4

## Explanation

HA   $\rightleftharpoons$    H+ + A$-$

Ka = ${{\left[ {{H^ + }} \right]\left[ {{A^ - }} \right]} \over {\left[ {HA} \right]}}$ = 10$-$5

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

$\Rightarrow $$\,\,\, 6 = - log [10-5] + log {{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}} \Rightarrow$$\,\,\,$ 6 = 5 + 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

$\therefore\,\,\,$ Salt : Acid = 10 : 1