Question

For a second-order reaction of the type rate=$k[A{]}^2$ the plot of $\frac{1}{[A{]}_1}$ versus t is linear with a:

• A. positive slope and zero intercept
• B. positive slope and non-zero intercept
• C. negative slope and zero intercept
• D. negative slope and non-zero intercept

Answer 1

The correct option is B positive slope and non-zero intercept

Given that for a second order reaction

$Rate=K[A{]}^2$

We know that rate of the reaction is decrease in the concentration of the reactant with respect to time.

$Rate=\frac{-d\left[A\right]}{dt}$

$\Rightarrow$ $\frac{-d\left[A\right]}{dt}=K[A{]}^2$

$\Rightarrow \frac{-d\left[A\right]}{[A{]}^2}=+Kdt$

Integrating both sides we have:-

$\Rightarrow -\int \frac{d\left[A\right]}{[A{]}^2}=+\int Kdt$

$\Rightarrow -\left[\frac{-1}{\left[A\right]}\right]=+Kt+C$

$\Rightarrow Kt+C=\frac{1}{\left[A\right]}$ $-\left(i\right)$

Now, from the equation $\left(i\right)$ we can see that graph of $\frac{1}{\left[A\right]}$ $V/s$ $t$ has a positive slope $K$ and non-zero intercept $C$.