Question

The rate of the reaction, \[\ce{2NO + O2 -> 2NO2}\], at 25°C is 0.028 mol L⁻¹ s⁻¹. The experimental rate is given by r = k [NO]² [O₂].

If the initial concentration of the reactants are [O₂] = 0.040 mol L⁻¹ and [NO] = 0.010 mol L⁻¹, the rate constant of the reaction is ______.

Options

• 7.0 × 10⁻² mol⁻¹ L s⁻¹

• 7.0 × 10⁻⁴ mol⁻² L² s⁻¹

• 7.0 × 10² mol⁻² L² s⁻¹

• 7.0 × 10³ mol⁻² L² s⁻¹

Answer 1

The rate of the reaction, \[\ce{2NO + O2 -> 2NO2}\], at 25°C is 0.028 mol L⁻¹ s⁻¹. The experimental rate is given by r = k [NO]² [O₂].

If the initial concentration of the reactants are [O₂] = 0.040 mol L⁻¹ and [NO] = 0.010 mol L⁻¹, the rate constant of the reaction is 7.0 × 10³ mol⁻² L² s⁻¹.

Explanation:

Given: Rate r = 0.028 mol L⁻¹ s⁻¹

[NO] = 0.010 mol L⁻¹

[O₂] = 0.040 mol L⁻¹

The given reaction is \[\ce{2NO + O2 -> 2NO2}\]

The rate law r = k [NO]² [O₂]

⇒ \[\ce{k = \frac{r}{[NO]^2 [O_2]}}\]

= \[\ce{\frac{0.028}{(0.010)^2 \times 0.040}}\]

= \[\ce{\frac{0.028}{4.0 \times 10^{-6}}}\]

= 7000 mol⁻² L² s⁻¹

= 7.0 × 10³ mol⁻² L² s⁻¹