Question

Consider this reaction:

Answer 1

Answer:

$C_{HI}=0.164M$

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:

$\frac{dC_{HI}}{dt} =-kC_{HI}$

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.