When $\text{ClNO}_2$ reacts with NO, the system reaches equilibrium:
$\text{ClNO}_2(g) + \text{NO}(g) \rightleftharpoons \text{NO}_2(g) + \text{ClNO}(g)$
In every $\text{ClNO}_2$ molecule, the nitrogen atom is linked to the chlorine atom and is also linked separately to each oxygen atom.
(a)[2]
Draw a ‘dot-and-cross’ diagram that represents the $\text{ClNO}_2$ molecule.
(b(i))[1]
Write the rate equation that applies to this reaction.
(b(ii))[1]
Deduce the units of the rate constant, $k$, when both gas concentrations are given in $\text{mol dm}^{-3}$ and the rate is given in $\text{mol dm}^{-3}\ \text{s}^{-1}$.
(b(iii))[1]
State and explain whether or not the reaction could take place in a single step.
(c(i))[2]
Draw a graph on the grid to show how the concentration of $\text{ClNO}_2$ changes over the first $0.20\ \text{s}$.
(c(ii))[1]
Deduce the concentration of the $\text{NO}_2$ product at $0.20\ \text{s}$.
(c(iii))[1]
After $20\ \text{s}$ the concentration of $\text{ClNO}_2$ remains constant. Explain this observation.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Eight electrons surrounding the N atom (N=O, N-O, N-Cl with N=O as dative)” …