Question

The reaction $2N_2{O}_5\left(g\right)\to 4N{O}_2\left(g\right)+O_2\left(g\right)$ is first order w.r.t. $N_2{O}_5$. Which of the following graphs would yield a straight line ?

• A. $log{p}_{N_2{O}_5}$ vs time with -ve slope
• B. $(p_{N_2{O}_5}{)}^{-1}$ vs time
• C. $p_{N_2{O}_5}$ vs time
• D. $log{p}_{N_2{O}_5}$ vs time with +ve slope

Answer 1

Answer: (A)

$log{p}_{N_2{O}_5}$ vs time with -ve slope

For a first order reaction, the graph of logarithm of the partial pressure of reactant to the time is a straight line with negative slope.

Hence, $log{p}_{N_2{O}_5}$ vs time t will give a straight line.