The graph obtained between \ln k ( k= Rate constant) on {\y}-axis and 1 \/ {\T} on {\x}-axis is a straight line. The slope of it is -4 \times 10^4 k. The activation energy of the reaction (in {\kJ} {\mol}^{-1} ) is ( {\R}=8.3 {\J} {\K}^{-1} {\mol}^{-1}) [TG EAPCET 12 May 2023 Shift 2]

TS EAMCET Engineering 12 May 2023 Shift 2 Solved Paper

Section: Chemistry

Question : 142 of 160

Marks: +1, -0

The graph obtained between ln‌k ( k= Rate constant) on y-axis and 1∕T on x-axis is a straight line. The slope of it is −4×104k. The activation energy of the reaction (in kJ‌mol−1 ) is
(R=8.3JK−1mol−1)

[TG EAPCET 12 May 2023 Shift 2]

Validate

Go to Question:

Prev Question

Next Question