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The rate constant for the first-order decomposition of C4H8 at 500oC is 9.2 x 10-3 s-1. How long will it take for 10.0% of a 0.100 M sample of C4H8 to decompose at 500oC

User Raz Ronen
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To calculate the time it will take for 10.0% of a 0.100 M sample of C4H8 to decompose at 500oC, we can use the first-order rate equation:

ln([C4H8]t/[C4H8]0) = -kt

where [C4H8]t is the concentration of C4H8 at time t, [C4H8]0 is the initial concentration, k is the rate constant, and t is the time.

Rearranging this equation, we get:

t = (ln([C4H8]0/[C4H8]t))/k

We know that k = 9.2 x 10-3 s-1, and we want to find the time it takes for 10.0% of the sample to decompose, which means [C4H8]t = 0.100 M x 0.10 = 0.010 M. Therefore, we can plug in these values and solve for t:

t = (ln(0.100/0.010))/9.2 x 10-3 s-1

t = 6.4 x 103 s

Therefore, it will take approximately 6.4 x 103 seconds, or 1.8 hours, for 10.0% of the sample to decompose at 500oC.

User Dhaivat Pandya
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