For the gas phase decomposition of cyclobutane,

\[ \text{(CH}_2\text{)}_4 \rightarrow 2\text{C}_2\text{H}_4 \]

the rate constant has been determined at several temperatures. When $\ln k$ in $\text{s}^{-1}$ is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of $-3.15 \times 10^4 \, \text{K}$ and a y-intercept of 35.9.

What is the activation energy for the gas phase decomposition of cyclobutane in $\text{kJ/mol}$?

Comparing this equation with the equation of a straight line, y = mx + b, we can see that ln(k) is the y-axis variable, 1/T is the x-axis variable, and (-Ea / (R)) is the slope of the line. ln(A) is the y-intercept.

Therefore, when ln(k) is plotted against 1/T, the resulting graph will be a straight line. The slope of the line will be (-Ea / (R)), and the y-intercept will be ln(A).By measuring the slope of the line and using the value of the gas constant (R), we can calculate the activation energy (Ea). Additionally, the y-intercept provides the value of ln(A), allowing us to determine the pre-exponential factor (A) by taking the expoential of ln(A).

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