Chemical Kinetics · Chemistry · TS EAMCET
MCQ (Single Correct Answer)
For the gaseous reaction, $\mathrm{N}_{2} \mathrm{O}_{5} \longrightarrow 2 \mathrm{NO}_{2}+\frac{1}{2} \mathrm{O}_{2}$
the rate can be expressed as
$ \begin{array}{l} -\frac{d\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right]}{d t}=K_{1}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \\\\ +\frac{d\left[\mathrm{NO}_{2}\right]}{d t}=K_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \\\\ +\frac{d\left[\mathrm{O}_{2}\right]}{d t}=K_{3}\left[\mathrm{~N}_{2} \mathrm{O}_{5}\right] \end{array} $
The correct relation between $K_{1}, K_{2}$ and $K_{3}$
| [A₂]₀ | [B₂]₀ | Initial rate of formation of AB (mol L⁻¹ s⁻¹) |
| 0.1 M | 0.1 M | 2.5 × 10⁻⁴ |
| 0.2 M | 0.1 M | 5.0 × 10⁻⁴ |
| 0.2 M | 0.1 M | 1.0 × 10⁻³ |
The graph obtained between $\ln k$ ( $k=$ rate constant) on $y$-axis and $1 / T$ on $x$-axis is a straight line. The slope of it is $-4 \times 10^4 \mathrm{~K}$. The activation energy of the reaction (in $\left.\mathrm{kJ} \mathrm{mol}^{-1}\right)$ is $\left(R=831 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\right)$