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Q.

Figure shows a graph in log⁡10 K vs 1T where K is rate constant and T is temperature. The straight line BC has slope, |tan⁡θ|=12.303 and an intercept of 5 on y-axis. Thus EA, the energy of activation is:

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a

2.303×2 cal

b

2/2.303 cal

c

2 cal

d

2.3032cal

answer is C.

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Detailed Solution

log⁡10 K vs. 1T is linear; slope=−EA2.303R intercept = log10ASince, |tan⁡θ|=12.303∴ EA2.303R=12.303⇒Ea=R=2cal
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