Question

For the reaction $N_2O_5 \longrightarrow 2NO_2 + \tfrac{1{2}O_2$, if} \[ -\frac{d[N_2O_5]}{dt} = K'[N_2O_5], \quad \frac{d[NO_2]}{dt} = K''[N_2O_5], \quad \frac{d[O_2]}{dt} = K'''[N_2O_5] \] then establish the relation between $K'$, $K''$, and $K'''$.

Hint:

Always divide the rate of change of concentration by the stoichiometric coefficient to establish relations between rate constants.

Answer 1

Step 1: Write the balanced chemical reaction.
\[ N_2O_5 \longrightarrow 2NO_2 + \tfrac{1}{2} O_2 \] Step 2: Define the rate of reaction.
The rate of reaction can be expressed as: \[ -\frac{1}{1}\frac{d[N_2O_5]}{dt} = \frac{1}{2}\frac{d[NO_2]}{dt} = \frac{1}{\tfrac{1}{2}} \frac{d[O_2]}{dt} \] \[ -\frac{d[N_2O_5]}{dt} = \frac{1}{2}\frac{d[NO_2]}{dt} = 2\frac{d[O_2]}{dt} \] Step 3: Substitute the given rate expressions.
\[ K'[N_2O_5] = \frac{1}{2}K''[N_2O_5] = 2K'''[N_2O_5] \] Step 4: Simplify the relations.
\[ K'' = 2K', \quad K''' = \tfrac{1}{2}K' \] Conclusion:
The relation between the constants is: \[ \boxed{K'' = 2K' \quad \text{and} \quad K''' = \tfrac{1}{2}K'} \]