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The correct nature of plot for first order reaction is (are):A. B. C. D. |
Answer» Correct Answer - A::D `k = (2.303)/(t)log.(c_(0))/(c )` `(k.t)/(2.303) = log c_(0) - log c` `log c = (-(k)/(2.303))t + log c_(0)` Thus, plot of `log c` vs `t` is a straight with negative slope. `t_(1//2) = x` should remain constant. `-(dc)/(dt) = k.c`, `log(-(dc)/(dt)) = log k + log c` So, the plot of `log(-(dc)/(dt))` vs `log c` should be a straight line of the slope equal to unity. |
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