Question

Consider this reaction:

2HI(g) → H2(g)+ I2(g)

At a certain temperature it obeys this rate law.

Rate= 8.74 x 10^-4 s^1

Suppose a vessel contains HI at a concentration of 0.330M. Calculate the concentration of HI in the vessel 800 seconds later. You may assume no other reaction is important. Round your answer to significant digit

Answer 1

`C_(HI)=0.164M`

Step-by-step explanation:

Hello,

In this case, based on the given units of the rate constant, it is a first-order reaction with respect to HI, therefore, the rate law is:

`(dC_(HI))/(dt) =-kC_(HI)`

Whose integration turns out:

`ln(C_(HI))=ln(C_(HI)^0)-kt\\C_(HI)=C_(HI)^0exp(-kt)`

Therefore, the concentration of HI after 800 second finally result:

`C_(HI)=0.330Mexp(-8.74x10^(-4)s^(-1)*800s)\\C_(HI)=0.164M`

Best regards.