Chapter 58: Problem 7

Recall that for a first-order reaction: $$ \ln (\mathrm{R})=\ln (\mathrm{R})_{0}-k t $$ a) When $t=t_{1 / 2}$, what is the value of $(\mathrm{R})$ in terms of $(\mathrm{R})_{0}$ ? b) Show that $t_{1 / 2}=\frac{\ln 2}{k}=\frac{0.693}{k}$ for a first-order reaction.

Short Answer

Expert verified

a) When $t = t_{1/2}$, $ R = \frac{R_0}{2}$. b) The half-life time formula for a first-order reaction is $ t_{1/2} = \frac{0.693}{k} $.

Step by step solution

Determine the Value of (R) at $t = t_{1/2}$

We take the formula for the first order reaction given by $ \ln(R) = \ln(R_0) - kt $ and we place $ t = t_{1/2} $ into the formula. We get: $ \ln(R) = \ln(R_0) - k t_{1/2} $. Given that $t_{1/2}$ signifies the half-life time, at this moment the amount of the reactant has been reduced by half i.e. it's half of the original amount. So, we may rewrite $R$ as $R_0/2$. Replacing $R$ in the equation we get: $ \ln \left(\frac{R_0}{2}\right) = \ln(R_0) - k t_{1/2} $. This is the answer for the first part.

Prove the Half-life Time Formula

To derive the formula for the half-life time of a first order reaction, we use the equation obtained in step 1: $ \ln \left(\frac{R_0}{2}\right) = \ln(R_0) - k t_{1/2} $. We perform a manipulation using the property of logarithms $\ln(a) - \ln(b) = \ln(a/b)$ and gets: $ \ln \left(\frac{R_0}{2}\right) - \ln(R_0) = - k t_{1/2} $. This can be rewritten as $ \ln \left(\frac{2}{1}\right) = k t_{1/2} $. The natural logarithm of 2 is 0.693. So, $ 0.693 = k t_{1/2} $. Finally we write down the half-life formula for the first-order reaction as $ t_{1/2} = \frac{0.693}{k} $.

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Recall that for a first-order reaction: $$ \ln (\mathrm{R})=\ln (\mathrm{R})_{0}-k t $$ a) When \(t=t_{1 / 2}\), what is the value of \((\mathrm{R})\) in terms of \((\mathrm{R})_{0}\) ? b) Show that \(t_{1 / 2}=\frac{\ln 2}{k}=\frac{0.693}{k}\) for a first-order reaction.

Short Answer

Step by step solution

Determine the Value of (R) at \(t = t_{1/2}\)

Prove the Half-life Time Formula

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