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प्रश्न
(b) Rate constant ‘k’ of a reaction varies with temperature ‘T’ according to the equation:
`logk=logA-E_a/2.303R(1/T)`
Where Ea is the activation energy. When a graph is plotted for `logk Vs. 1/T` a straight line with a slope of −4250 K is obtained. Calculate ‘Ea’ for the reaction.(R = 8.314 JK−1 mol−1)
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उत्तर
`log k=logA-E_a/2.303R(1/T)`
Ea → Activation energy
The above equation is like y = mx + c where if we plot y v/s x we get a straight line with slope ‘m’ and intercept ‘c’.
So, slope is equal to`=-E_a/2.303R`
`-E_a/2.303R=-4250k=>E_a=4250 xx 2.303 xx 8.314 =81,375.3535 j mol^(-1)`
`E_a=81.3753 KJ mol^(-1)`
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