At 298 K,for a first order reaction (A right | vedclass

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(D) For a first order reaction,the integrated rate equation is: $\ln(a-x) = -Kt + \ln a$. Comparing this with the equation of a straight line $y = mx

Vedclass · Accepted Answer

$10^{-2}; 10^{-1}$

Vedclass · Answer

(D) For a first order reaction,the integrated rate equation is: $\ln(a-x) = -Kt + \ln a$. Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln(a-x)$,$x = 	ext{time}$,slope $m = -K$,and intercept $c = \ln a$. From the given graph: Intercept $c = -2.303 = \ln a$. Therefore,$a = e^{-2.303} \approx 10^{-1} \ mol \ L^{-1}$. Slope $m = -(10)^{-2} = -K$. Therefore,$K = 10^{-2} \ s^{-1}$. Thus,the rate constant is $10^{-2} \ s^{-1}$ and the initial concentration is $10^{-1} \ mol \ L^{-1}$.

At $298 \ K$,for a first order reaction $(A \rightarrow P)$ the following graph is obtained. The rate constant (in $s^{-1}$) and initial concentration (in $mol \ L^{-1}$) of '$A$' are respectively ($y$-axis $= \ln(a-x)$; $x$-axis $=$ time in sec).

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