The decomposition of ethane, C2H6, is a first-order reaction. It is found that it takes 212 seconds to decompose 0.00839 M C2H6 to 0.00768 M. What is the rate of decomposition (in mol/L-h) when [C2H6] = 0.00422 M?
A.
6.43 x 10^-3 mol/L-h
B.
6.34 x 10^-5 mol/L-h
C.
6.34 x 10^-3 mol/L-h
D.
6.304 x 10^-3 mol/L-h
To find the rate of decomposition, we can use the first-order rate law equation:
rate = k[A]
where rate is the rate of decomposition, k is the rate constant, and [A] is the concentration of the reactant.
Given that the decomposition of ethane is a first-order reaction, the rate law equation simplifies to:
rate = k[C2H6]
Now, let's calculate the rate constant (k) using the given data. We have:
t = 212 seconds
[C2H6]0 = 0.00839 M
[C2H6]t = 0.00768 M
Since it is a first-order reaction, we can use the integrated rate law equation for a first-order reaction:
ln([C2H6]0/[C2H6]t) = kt
Rearranging the equation:
k = ln([C2H6]0/[C2H6]t) / t
Substituting the given values:
k = ln(0.00839 M / 0.00768 M) / 212 seconds
Now, let's calculate k:
k = ln(1.0911458) / 212 seconds
k ≈ 0.0022894 seconds^-1
Now that we have the rate constant, we can calculate the rate of decomposition when [C2H6] = 0.00422 M. Using the rate law equation:
rate = k[C2H6]
Substituting the values:
rate = (0.0022894 seconds^-1)(0.00422 M)
rate ≈ 9.66 x 10^-6 mol/(L-seconds)
To convert to mol/L-hour, we can multiply by the conversion factor:
rate ≈ 9.66 x 10^-6 mol/(L-seconds) * (1 hour/3600 seconds)
rate ≈ 2.68 x 10^-9 mol/(L-hour)
Therefore, the rate of decomposition when [C2H6] = 0.00422 M is approximately 2.68 x 10^-9 mol/L-hour.
None of the provided answer choices match this value, so there may be an error in the question or answer choices.