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CBSE Class 12 Chemistry Solved Question Paper 2018

Short Answer Type

21.

A first-order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.

(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^–1 mol^–1)

22.

What happens when

A freshly prepared precipitate of Fe(OH)₃ is shaken with a small amount of FeCl₃ solution?

23.

What happens when

Persistent dialysis of a colloidal solution is carried out?

24.

What happens when

An emulsion is centrifuged?

25.
For the reaction
$2 A g C l (s) + H_{2} (g) (1 a t m) \overset{}{\to} 2 A g (s) + 2 H^{+} (0.1 M) + 2 C l^{-} (0.1 M), ∆ G^{o} = - 43600 J a t 25^{o} C$
Calculate the e.m.f. of the cell.
[log 10^–n = – n]

$G i v e n : 2 A g C l (s) + H_{2} (g) (1 a t m) \overset{}{\to} 2 A g (s) + 2 H^{+} + 2 C l^{-} ∆ G^{o} = - 43600 J T = 25^{o} C = 298 K W e k n o w, ∆ G^{o} = - n F E_{c e l l}^{°} n = 2 ∴ E_{c e l l}^{°} = \frac{∆ G^{o}}{n F} = \frac{- (- 43600)}{2 x 96500} = 0.226 V E_{c e l l} = E_{c e l l}^{o} - \frac{0.059}{n} \log \frac{[H^{+}]^{2} [C l^{-}]^{2} [A g]}{[A g C l]^{2} [H_{2}]} = 0.226 - \frac{0.059}{2} \log {[H^{+}]}^{2} {[C l^{-}]}^{2} = 0.226 - \frac{0.059}{2} \log (0.1)^{2} (0.1)^{2} = 0.226 - \frac{0.059}{2} \log (0.1 x 0.1 x 0.1 x 0.1) = 0.226 - \frac{0.059}{2} \log 10^{- 4} = 0.226 - \frac{0.059}{2} (- 4) = 0.226 + 0.118 E_{c e l l} = 0.344 V$