Question

1. nitric oxide, no_(g), reacts with chlorine gas, cl_(2(g)), in the balanced chemical equation below: 2no_(g)+cl_(2(g))→2nocl_(g) a chemical technician carries out this reaction three times at the same temperature, and collects the data seen below. a) determine the rate law (leaving the constant as \k\) and indicate your reasoning. b) determine the value of \k\ (rate constant) - you could try the units if you wish for a bonus mark. c) according to these results, what would the rate be if the concentrations of the reactants are both 0.4 mol/l?

Explanation:

Part (a)

Step 1: Assume rate law form

Let the rate law be \( \text{Rate} = k[\text{NO}]^m[\text{Cl}_2]^n \), where \( m \) and \( n \) are the orders with respect to \( \text{NO} \) and \( \text{Cl}_2 \) respectively.

Step 2: Find order with respect to \( \text{Cl}_2 \)

Compare trials 1 and 2 (where \( [\text{NO}] \) is constant).
Trial 1: \( [\text{Cl}_2] = 0.10 \, \text{mol/L} \), Rate \( = 1.8 \times 10^{-2} \, \text{mol/L·s} \)
Trial 2: \( [\text{Cl}_2] = 0.20 \, \text{mol/L} \), Rate \( = 3.6 \times 10^{-2} \, \text{mol/L·s} \)
The ratio of rates: \( \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{3.6 \times 10^{-2}}{1.8 \times 10^{-2}} = 2 \)
The ratio of \( [\text{Cl}_2] \): \( \frac{0.20}{0.10} = 2 \)
So, \( 2 = 2^n \implies n = 1 \) (order with respect to \( \text{Cl}_2 \) is 1).

Step 3: Find order with respect to \( \text{NO} \)

Compare trials 2 and 3 (where \( [\text{Cl}_2] \) is constant).
Trial 2: \( [\text{NO}] = 0.10 \, \text{mol/L} \), Rate \( = 3.6 \times 10^{-2} \, \text{mol/L·s} \)
Trial 3: \( [\text{NO}] = 0.20 \, \text{mol/L} \), Rate \( = 1.44 \times 10^{-1} \, \text{mol/L·s} \)
The ratio of rates: \( \frac{\text{Rate}_3}{\text{Rate}_2} = \frac{1.44 \times 10^{-1}}{3.6 \times 10^{-2}} = 4 \)
The ratio of \( [\text{NO}] \): \( \frac{0.20}{0.10} = 2 \)
So, \( 4 = 2^m \implies m = 2 \) (order with respect to \( \text{NO} \) is 2).

Step 4: Write rate law

Substituting \( m = 2 \) and \( n = 1 \) into the rate law: \( \text{Rate} = k[\text{NO}]^2[\text{Cl}_2] \)

Step 1: Use a trial to solve for \( k \)

Use trial 1: \( [\text{NO}] = 0.10 \, \text{mol/L} \), \( [\text{Cl}_2] = 0.10 \, \text{mol/L} \), Rate \( = 1.8 \times 10^{-2} \, \text{mol/L·s} \)
From rate law \( \text{Rate} = k[\text{NO}]^2[\text{Cl}_2] \), rearrange for \( k \):
\( k = \frac{\text{Rate}}{[\text{NO}]^2[\text{Cl}_2]} \)

Step 2: Substitute values

\( k = \frac{1.8 \times 10^{-2} \, \text{mol/L·s}}{(0.10 \, \text{mol/L})^2(0.10 \, \text{mol/L})} \)
Calculate denominator: \( (0.10)^2(0.10) = 0.001 \, \text{mol}^3/\text{L}^3 \)
\( k = \frac{1.8 \times 10^{-2}}{0.001} = 18 \, \text{L}^2/(\text{mol}^2·\text{s}) \)

Step 1: Use the rate law and \( k \)

Rate law: \( \text{Rate} = k[\text{NO}]^2[\text{Cl}_2] \), \( k = 18 \, \text{L}^2/(\text{mol}^2·\text{s}) \), \( [\text{NO}] = 0.4 \, \text{mol/L} \), \( [\text{Cl}_2] = 0.4 \, \text{mol/L} \)

Step 2: Substitute values

\( \text{Rate} = 18 \, \text{L}^2/(\text{mol}^2·\text{s}) \times (0.4 \, \text{mol/L})^2 \times (0.4 \, \text{mol/L}) \)
Calculate \( (0.4)^2(0.4) = 0.064 \, \text{mol}^3/\text{L}^3 \)
\( \text{Rate} = 18 \times 0.064 = 1.152 \, \text{mol/L·s} \) (or \( 1.152 \times 10^0 \, \text{mol/L·s} \))

Answer:

The rate law is \( \boldsymbol{\text{Rate} = k[\text{NO}]^2[\text{Cl}_2]} \). The order with respect to \( \text{NO} \) is 2 (from trials 2→3, doubling \( [\text{NO}] \) quadruples the rate) and with respect to \( \text{Cl}_2 \) is 1 (from trials 1→2, doubling \( [\text{Cl}_2] \) doubles the rate).

Part (b)