Question

(a )Derive relationship between Kp and Kc for the following reaction.
2SO3(g) ⇄ 2SO2+O2
(b)Calculate the pH of 0.3 g of Ca(OH)2 in 500ml of solution? [log 16.2=1.209]

Answer 1

Answer:

ANSWER

Initially, number of moles of SO

2

,O

2

andSO

3

are 2, 1.5 and 0 respectively.

2 moles of KMnO

4

reacts with 5 moles of SO

2

.

0.4 moles of KMnO

4

will react with 1 moles of SO

2

.

Thus at equilibrium number of moles of SO

2

,O

2

andSO

3

are 1, 1 and 1 respectively.

The volume is 5 L.

Thus the equilibrium concentrations of SO

2

,O

2

andSO

3

are 0.2, 0.2 and 0.2 respectively.

The expression for the equilibrium constant is K

p

[SO

2

]

2

[O

2

]

[SO

3

]

2

=

0.2

2

×0.2

0.2

2

=5.

Answer 2

Given:

$2SO_{3}(g)$⇄ $2SO_{2}+O_{2}$.

To find:

Calculate the pH value and relationship between the Kp and Kc reaction.

Explanation:

Molarity of $CaOH_{2}$.

$=\frac{Mass of Ca(OH)_{2}/Molar Mass }{Volume of solution in litres}$

$=\frac{(0.3g)/74 g mol}{0.5 l}$

$=8.1*10^{-3}molL^{-1}$

$=8.1*10^{-3}M$

$Ca(OH)_{2}$→$Ca^{2+}(aq)+2OH^{-}(aq)$

$=8.1*10^{-3}M+16.2*10^{-3}M$

$pOH=-log[OH]^{-}$

$=-log(16.2*10^{-3})$

=$=-(log 16.2-3 log 10)$

$=(3-log16.2)$

$=(3-1.209)$

$=1.791$

$pH=14-pOH$

$=14-1.791$

$=12.21$

Answer:

Therefore, the pH value as 12.21.