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A graph plotted between log k versus 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 correct Answer is:
B
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The graph between log k versus 1//T is a straight line.

i) Write the mathematics expressions relating the variation of the rate constant of a reaction with temperatures. ii) How can you graphically find the activation energy of the reaction from the above expression? iii) The slope of the line in the graph of log k(k=rate constant) versus 1//T is -5841 . Calculate the activation energy of the reaction.

Knowledge Check

  • A graph is ploted between log K virsus (1)/(T) for calculation of activation energy (E_(a)) . The correct plot is

    A
    B
    C
    D

  • A graph plotted between log k vs (1)/(T) is represented by

    A
    B
    C
    D

  • The graph plotted between concentration versus time

    A
    It gives rate of disappearance of reactant
    B
    Rate `=(d[C_2-C_1)])/(t_2-t_1)`
    C
    Both (1) & (2)
    D
    It predicts the order of reaction

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    The slope of straight line graph between ln k us (1)/(T) is equal to 2.4 xx 10^(4)K . Calculate the activation energy of the reaction at 400 K. (K represents rate constant)

    If a graph is plotted between ln k and 1//T for the first order reaction, the slope of the straight line so obtained is given by

    A graph plotted between log t_(50%) vs log concentration in a straight line. What conclusion can you draw from this graph?

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