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Rate constant k of a reaction varies wit...

Rate constant k of a reaction varies with temperature according to the equation
logk = constant `-E_a/(2.303R)(1/T)`
where `E_a` is the energy of activation for the reaction . When a graph is plotted for log k versus `1/T` , a straight line with a slope - 6670 K is obtained. Calculate the energy of activation for this reaction

Text Solution

AI Generated Solution

To solve the problem, we will follow these steps: ### Step 1: Understand the given equation The relationship between the rate constant \( k \) and temperature \( T \) is given by: \[ \log k = \text{constant} - \frac{E_a}{2.303R} \left(\frac{1}{T}\right) \] where: ...
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Rate constant k for a reaction varies with temperature according to the equation log k ="constant"-(E_(a))/(2.303R)*(1)/(T) where E_(a) is the energy of activation for the reaction. When a graph is plotted for log k vs 1//T , a straight line with a slope of -6670 K is obtained. Calculate energy of activation for this reaction. State the units. " "[R=8.314JK^(-1)"mol"^(-1)]

Rate constant K of a reaction varies with temperature ‘T’ according to the equation : log K= log A- E_a/(2.303R)(1/T) where E_a is the activation energy. When a graph is plotted for log K vs 1/T, a straight line with a slope of − 4250 K is obtained. Calculate E_a for the reaction.

Knowledge Check

  • Rate constant k of a reaction varies with temperature according to the equation, log k = constant - (E_a)/(2.303RT) When a graph is plotted for logk versus 1/T a straight line with a slope -5632 is obtained. The energy of activation for this reaction is

    A
    127.67 kJ `mol^(-1)`
    B
    107.84 kJ `mol^(-1)`
    C
    86 kJ `mol^(-1)`
    D
    246.8 kJ `mol^(-1)`

  • A plot of k vs 1//T for a reaction gives the slope -1 xx 10^(4) K. The energy of activation for the reaction is:

    A
    8314 J `mol^(-1)`
    B
    `1.202 kJmol^(-1)`
    C
    `12.02 J mol^(-1)`
    D
    `83.14 kJ mol^(-1)`

  • A graph plotted between log k versus 1/T for calculating activation energy is shown by

    A
    B
    C
    D

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    The rate constant 'k'. For a reaction varies with temperature 'T' according to the question. log k = log A-(E_(a))/(2.303R)(1/T) Where E_(a) is the activation energy. When a graph is plotted for log k vs 1//T , a straight line with a slope of -4250 K is obtained. Calcualte E_(a) for this reaction.

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