The Arrhenius plots of two reactions,I and II | vedclass

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Question

(A) The Arrhenius equation is given by $k = A e^{-E_a / R T}$.Taking the natural logarithm on both sides,we get $\ln k = \ln A - \frac{E_a}{R} \left(

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$E_I > E_{II}$ and $A_I > A_{II}$

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(A) The Arrhenius equation is given by $k = A e^{-E_a / R T}$.Taking the natural logarithm on both sides,we get $\ln k = \ln A - \frac{E_a}{R} \left( \frac{1}{T} \right)$.Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln k$ and $x = 1/T$:The slope $m = -E_a / R$ and the intercept $c = \ln A$.From the graph,the line for reaction $I$ is steeper than the line for reaction $II$. Since the slope is negative,a steeper line indicates a larger magnitude of slope,meaning $E_I > E_{II}$.Also,the intercept on the $y$-axis (at $1/T = 0$) for line $I$ is higher than for line $II$. Since the intercept is $\ln A$,this implies $\ln A_I > \ln A_{II}$,which means $A_I > A_{II}$.Therefore,the correct option is $A$.

The Arrhenius plots of two reactions,$I$ and $II$ are shown graphically. The graph suggests that

Similar Questions

For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

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($k$ is the rate constant of an elementary reaction and $T$ is temperature in absolute scale)

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