Question

The initial concentration of NOCl in the second-order reaction 2NOClâ2NO+Cl2 is 0.878M. After 763,200 seconds, the concentration of NOCl is 0.83M. What is the rate constant k for the reaction? Report your answer in scientific notation rounded to two significant figures. Use the multiplication symbol when reporting your answer rather than the letter x. Provide your answer below: $$ 1/M s

Answer 1

`k=8.63x10^(-8)(1)/(M*s)`

Step-by-step explanation:

Hello!

In this case, since the differential rate law of a second-order reaction is:

`(dC_A)/(dt)=-kC_A^2`

Whereas A stands for NOCl and the corresponding integrated rate law is:

`(1)/(C_A) =kt+(1)/(C_A_0)`

Thus, since we know the concentrations and the elapsed time, we compute the rate constant as shown below:

`k=( (1)/(C_A)-(1)/(C_A_0) )/t\\\\k=( (1)/(0.83M)-(1)/(0.878M) )/763,200s\\\\k=8.63x10^(-8)(1)/(M*s)`

Best regards!