Question

Ammonium ion ($NH_4^+$) reacts with nitrite ion ($NO_2^-$). Based on the experimental data provided, which of the following is the rate law? ________.

• A. $Rate=k[NH_{4}^{+}]^{1/2}[NO_{2}^{-}]$
• B. $Rate=k[NH_{4}^{+}][NO_{2}^{-}]$
• C. $Rate=k[NH_{4}^{+}]^{0}[NO_{2}^{-}]$
• D. $Rate=k[NH_{4}^{+}][NO_{2}^{-}]^{1/2}$
• E. $Rate=k[NH_{4}^{+}][NO_{2}^{-}]^{2}$

Hint:

If doubling concentration doubles rate, order is 1. If doubling concentration quadruples rate, order is 2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
This problem requires determining the rate law of a reaction using the method of initial rates. The rate law expresses the reaction rate in terms of the concentration of the reactants. For a generic reaction \(aA + bB \to \text{products}\), the rate law has the form \(\text{Rate} = k[A]^x[B]^y\), where \(x\) and \(y\) are the reaction orders with respect to reactants A and B, which must be determined experimentally.
Step 2: Key Formula or Approach
Let the rate law be \(\text{Rate} = k[\text{NH}_4^+]^x[\text{NO}_2^-]^y\). We will find the orders \(x\) and \(y\) by comparing the initial rates from different experiments where the concentration of one reactant is varied while the other is held constant.
Step 3: Detailed Explanation
1. Determine the order with respect to [NH\(_4^+\)] (find x).
Compare Experiment I and Experiment II. In this pair, [\text{NO}\(_2^-\)] is constant (0.020 M), while [\text{NH}\(_4^+\)] changes from 0.010 M to 0.015 M.
The ratio of the rates is: \[ \frac{\text{Rate II}}{\text{Rate I}} = \frac{k(0.015)^x(0.020)^y}{k(0.010)^x(0.020)^y} = \frac{0.030}{0.020} \] The terms with \(k\) and \((0.020)^y\) cancel out. \[ \left(\frac{0.015}{0.010}\right)^x = \frac{3}{2} \] \[ (1.5)^x = 1.5 \] This implies that the order \(x = 1\). The reaction is first-order with respect to \(\text{NH}_4^+\).
2. Determine the order with respect to [NO\(_2^-\)] (find y).
Compare Experiment I and Experiment III. In this pair, [\text{NH}\(_4^+\)] is constant (0.010 M), while [\text{NO}\(_2^-\)] changes from 0.020 M to 0.010 M.
The ratio of the rates is: \[ \frac{\text{Rate I}}{\text{Rate III}} = \frac{k(0.010)^x(0.020)^y}{k(0.010)^x(0.010)^y} = \frac{0.020}{0.005} \] The terms with \(k\) and \((0.010)^x\) cancel out. \[ \left(\frac{0.020}{0.010}\right)^y = 4 \] \[ (2)^y = 4 \] This implies that the order \(y = 2\). The reaction is second-order with respect to \(\text{NO}_2^-\).
3. Write the complete rate law.
Combining the orders we found, the rate law is: \[ \text{Rate} = k[\text{NH}_4^+]^1[\text{NO}_2^-]^2 \] This matches option (E).
Step 4: Final Answer
The rate law for this reaction is Rate = k[\text{NH}\(_4^+\)][\text{NO}\(_2^-\)]\(^2\).