Question

5. The reaction, SO2C2-SO2 + Cl2, is a first-order gas reaction with k = 2.2 x 10-s.
at 320°C. What percent SO2Cl2 is decomposed on heating at 320°C for 90 min?​

Answer 1

Answer:

94.96%.

Explanation:

The time taken will be 90 x 60=5400 seconds. While for the first order reaction we will  get that.

The reaction is SO2C2 -> SO2 + Cl2.

ln{[SOCl2]/[SO2][Cl2]}/t=k Where k is the reaction rate and t is the time taken.

So, ln{[SOCl2]/[SO2][Cl2]}=2.2 x 10^-5 x 5400 which will be 0.119.  

log{[SOCl2]/[SOCl2]} = 0.119/2.303 = 0.0516.  

Therefore, the value will be e^0.0516 =1.053 =[SOCl2]/[SO2][Cl2] which on solving we will get that the percentage of decomposition will be. If we let that the final reactant as 100. So, the product concentration or SO2Cl2 will be           100/1.053 = 94.96%.