Question

The decomposition of a compound in a first-order solution, with
an activation energy of 52.3 kJ mol-1. A solution with 10% A
decomposes 10% in 10 min at 10 ° C. What percentage of
decomposition can be found with a 20% solution after 20 min at 20 °
C? Answer: *36.21%*

Answer #1

decomposition of a compound in a first-order solution

A ? B

Given that

10% A decomposes 10% in 10 minute at 10° C

Initial amount of A

a = 10

Amount left after time t = 10 min

(a - x) = 10 * (90/100) = 9

Rate constant at 10°C

K_{10} = (2.303/t) * log_{10}(a/a-x)

= (2.303/10) * log_{10}(10/9)

= 0.0105

Now calculate the rate constant at 20°C

2.303 *
log_{10}(K_{2}_{0}/K_{10}) =
(Ea/R)[(T_{2} –
T_{1})/T_{1}T_{2}]

2.303 *
log_{10}(K_{2}_{0}/K_{10}) = {(52.3
* 10^{3})/8.314}[(293 - 283)/(293 * 283]

(K_{2}_{0}/K_{10}) = 2.135

K_{2}_{0} = 2.135 * 0.0105 = 0.0224175

Let amount left 20% solution after 20 minute at 20° C = z

K_{2}_{0} = (2.303/t) * log_{10}{a/(a –
x)}

0.0224175 = (2.303/20) * log_{10}(20/z)

0.19468 = log_{10}(20/z)

20/z = 1.5655

z = 12.775

% decomposition = (a – z)*100/a

= (20 – 12.775)*100/20

= 36.125%

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