Problem: Consider the following reaction: 2O3(g) → 3O2(g). The rate law for this reaction is as follows: Rate = k[O3]2/[O2]Suppose that a 1.0-L reaction vessel initially contains 1.0 mol of O3 and 1.0 mol of O2. What fraction of the O3 will have reacted when the rate falls to one-half of its initial value?

🤓 Based on our data, we think this question is relevant for Professor Bueno's class at UMD.

FREE Expert Solution

We are asked to determine what fraction of the O₃ will have reacted when the rate falls to one-half of its initial value.

Recall that the rate law only focuses on the reactant concentrations and has a general form of:

$\boxed{\mathbf{rate}\mathbf{ }\mathbf{law}\mathbf{=}\mathbf{k}{\left[\mathbf{A}\right]}^{\mathbf{x}}{\left[\mathbf{B}\right]}^{\mathbf{y}}}$

k = rate constant
A & B = reactants
x & y = reactant orders

Given:            

$\boxed{\mathbf{Rate}\mathbf{=}\mathbf{k}\frac{{\mathbf{\left[}{\mathbf{O}}_{\mathbf{3}}\mathbf{\right]}}^{\mathbf{2}}}{\mathbf{\left[}{\mathbf{O}}_{\mathbf{2}}\mathbf{\right]}}}$

[O₃] = 1.0 mol

[O₂] = 1.0 mol

Volume of Container = 1 L