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Step-by-Step Solution
Step 1: Write the Rate Equation for a First-Order Reaction
For a first-order reaction, the rate constant k is given by the formula:
k = \frac{1}{t}\,\ln\!\Bigl(\frac{a}{a - x}\Bigr),
where
a is the initial amount (or concentration) of the reactant,
x is the amount (or concentration) of the reactant decomposed after time t .
Rearranging for time t , we get:
t = \frac{1}{k}\,\ln\!\Bigl(\frac{a}{a - x}\Bigr).
Step 2: Apply the Formula for 60% Decomposition
When the reaction is 60% complete:
x = 0.6\,a,
Time taken t_1 = 540\,\mathrm{s}.
Substitute into the first-order expression to find k (the rate constant):
k = \frac{1}{t_1} \,\ln\!\Bigl(\frac{a}{a - 0.6a}\Bigr)
= \frac{1}{540} \,\ln\!\Bigl(\frac{a}{0.4\,a}\Bigr).
Since \frac{a}{0.4\,a} = \frac{1}{0.4} = 2.5 ,
k = \frac{1}{540} \,\ln(2.5).
(Note that we will mainly need the ratio form of the times; hence an exact numeric value of k may not be strictly necessary.)
Step 3: Express the Time for 90% Decomposition
When the reaction is 90% complete:
x = 0.9\,a,
The required time is t_2 (which we want to determine).
From the first-order expression:
t_2 = \frac{1}{k} \,\ln\!\Bigl(\frac{a}{a - 0.9a}\Bigr)
= \frac{1}{k} \,\ln\!\Bigl(\frac{a}{0.1\,a}\Bigr).
Since \frac{a}{0.1\,a} = \frac{1}{0.1} = 10,
t_2 = \frac{1}{k}\,\ln(10).
Step 4: Relate t_2 to t_1 Using the Ratio of Logarithms
Recall from Step 2:
t_1 = \frac{1}{k}\,\ln\!\Bigl(\frac{1}{0.4}\Bigr),
\quad
t_2 = \frac{1}{k}\,\ln(10).
Taking the ratio:
\frac{t_1}{t_2}
= \frac{\ln\!\bigl(\frac{1}{0.4}\bigr)}{\ln(10)}.
We know t_1 = 540\,\mathrm{s}, thus
\frac{540}{t_2}
= \frac{\ln\!\bigl(\frac{1}{0.4}\bigr)}{\ln(10)}.
Step 5: Plug in Numerical Values and Solve for t_2
Given:
\ln(10) = 2.3,
\log_{10}(2) = 0.3 \quad(\text{hence}~\log_{10}(4) = 0.6 \text{ and } \ln(4) \approx 2.3 \times 0.6 = 1.38),
We have
\ln\!\bigl(\tfrac{1}{0.4}\bigr) = \ln(2.5) = \ln\!\bigl(\tfrac{5}{2}\bigr)
\approx \ln(5) - \ln(2).
\quad
(\text{Detailed values may vary, but we can keep it symbolic for the ratio approach.})
In the provided solution approach,
\frac{540}{t_2} = \frac{\ln\!\bigl(\tfrac{1}{0.4}\bigr)}{\ln(10)}
\approx \frac{2.3 - 2.3 \times 0.6}{2.3} = 0.4,
because \ln(4) = 2.3 \cdot 0.6 was used to simplify terms. Therefore,
\frac{540}{t_2} = 0.4
\quad \Longrightarrow \quad
t_2 = \frac{540}{0.4} = 1350\,\mathrm{s}.
Answer
The time required for 90% decomposition is
t_2 = 1350\,\mathrm{s}.
