The slope of the line in the graph of log k(k=rate constant) vs 1T for a reaction is −5841K. Calculate the energy of activation for this reaction R=8.314JKmol−

Slope =−Ea2.303R∴ −5841K=−Ea2.303×8.314JK−mol−∴ Ea=11.18×104Jmol−.

The slope of the line in the graph of log k(k=rate constant) vs 1T for a reaction is −5841K. Calculate the energy of activation for this reaction R=8.314JKmol−

Slope =−Ea2.303R∴ −5841K=−Ea2.303×8.314JK−mol−∴ Ea=11.18×104Jmol−.

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The slope of the line in the graph of log k(k=rate constant) vs $\frac{1}{T}$ for a reaction is $- 5841 K .$ Calculate the energy of activation for this reaction $(R = 8 .314 {JKmol}^{-})$

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11.18x10⁴Jmol^-1

11.18x10⁵Jmol^-1

111.8Jmol^-1

1.118x10⁵kJmol^-1

answer is A.

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Slope $= \frac{- E_{a}}{2.303 R}$

$\begin{array}{l} ∴ - 5841 K = \frac{- E_{a}}{2.303 \times 8.314 {JK}^{-} {mol}^{-}} \\ ∴ E_{a} = 11.18 \times 10^{4} {Jmol}^{-} . \end{array}$

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The slope of the line in the graph of log k(k=rate constant) vs 1T for a reaction is −5841K. Calculate the energy of activation for this reaction R=8.314JKmol−

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The slope of the line in the graph of log k(k=rate constant) vs $\frac{1}{T}$ for a reaction is $- 5841 K .$ Calculate the energy of activation for this reaction $(R = 8 .314 {JKmol}^{-})$