For the reaction 2N_2O_{5(g)} rightarrow 4NO_ | vedclass

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(A) For a first-order reaction,the integrated rate equation is given by: $\ln(P_t) = \ln(P_0) - kt$. Converting this to base $10$ logarithm: $\log(P_t

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$\log(P_{N_2O_5})$ versus time with a negative slope

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(A) For a first-order reaction,the integrated rate equation is given by: $\ln(P_t) = \ln(P_0) - kt$. Converting this to base $10$ logarithm: $\log(P_t) = \log(P_0) - \frac{kt}{2.303}$. Comparing this with the equation of a straight line $y = mx + c$,where $y = \log(P_t)$,$x = t$,$m = -\frac{k}{2.303}$ (slope),and $c = \log(P_0)$ (intercept). Thus,a plot of $\log(P_{N_2O_5})$ versus time $t$ yields a straight line with a negative slope.

For the reaction $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$,which is a first-order reaction with respect to $N_2O_5$,which of the following plots gives a straight line?

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