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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

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sebassandin

Answer:

[tex]C_{HI}=0.164M[/tex]

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:

[tex]\frac{dC_{HI}}{dt} =-kC_{HI}[/tex]

Whose integration turns out:

[tex]ln(C_{HI})=ln(C_{HI}^0)-kt\\C_{HI}=C_{HI}^0exp(-kt)[/tex]

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

[tex]C_{HI}=0.330Mexp(-8.74x10^{-4}s^{-1}*800s)\\C_{HI}=0.164M[/tex]

Best regards.

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