anonymous
  • anonymous
The decomposition of SOCl2 is first-order in SOCl2. If the half-life for the reaction is 4.1 hr, how long would it take for the concentration of SOCl2 to drop from 0.36M to 0.045M? Please help explain half-life and how to do the problem.
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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aaronq
  • aaronq
for a first order reaction, you can use: \(t_{1/2}=\dfrac{ln2}{k}\), to find k, the decay constant. After you can use the exponential growth/decay equation: \(A_{t}=A_0*e^{-kt}\) where \(A_{t}\) is the amount after t time elapsed, \(A_0\) is the initial amount. t is time, and k is the decay constant (you found above).
anonymous
  • anonymous
A0 is 0.36
aaronq
  • aaronq
yep

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anonymous
  • anonymous
A(t) is 0.045M
aaronq
  • aaronq
correct
anonymous
  • anonymous
So the 4.1 hr would be the t right? But what I'm confuse about is the rate constant (-k) and the e?
aaronq
  • aaronq
e is eulers number, and the k is the constant you found using the first equation by plugging in the half life (\(t_{1/2}\)). t is what you're solving for
anonymous
  • anonymous
I never did the first equation to this problem.
anonymous
  • anonymous
I think I figured it out. 4.1h is equal to 14760s. Since k equals the rate reaction is 4.97E-5. By taking the In 0.045/0.36 which equals -4.97E-5 we get 2.08. 2.08 divide into 4.97E-5 which is 41839.9s equals t.
anonymous
  • anonymous
Thanks for the help.

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