Question

Ammonium ion (NH\textsubscript{4}\textsuperscript{+}) reacts with nitrite ion (NO\textsubscript{2}\textsuperscript{--}) in aqueous solution according to the equation
NH\textsubscript{4\textsuperscript{+}(aq) + NO\textsubscript{2}\textsuperscript{--}(aq) \(\rightarrow\) N\textsubscript{2}(g) + 2H\textsubscript{2}O(l)

• A. rate = \(k\) [NH\textsubscript{4}\textsuperscript{+}] [NO\textsubscript{2}\textsuperscript{--}]\textsuperscript{4}
• B. rate = \(k\) [NH\textsubscript{4}\textsuperscript{+}] [NO\textsubscript{2}\textsuperscript{--}]
• C. rate = \(k\) [NH\textsubscript{4}\textsuperscript{+}] [NO\textsubscript{2}\textsuperscript{--}]\textsuperscript{2}
• D. rate = \(k\) [NH\textsubscript{4}\textsuperscript{+}]\textsuperscript{2} [NO\textsubscript{2}\textsuperscript{--}]
• E. rate = \(k\) [NH\textsubscript{4}\textsuperscript{+}]\textsuperscript{1/2} [NO\textsubscript{2}\textsuperscript{--}]\textsuperscript{1/4}

Hint:

If doubling the concentration doubles the rate, the order is 1. If doubling the concentration quadruples the rate, the order is 2. Quick mental ratios are the key to solving these table-based questions in seconds!

Answer 1

The Correct Option is C

Concept: The rate law expresses the relationship between the reaction rate and reactant concentrations. The orders (exponents) must be determined experimentally.
• General Rate Law: Rate = \(k [NH_4^+]^x [NO_2^-]^y\).
• Method of Initial Rates: By keeping one concentration constant and varying the other, we can observe the effect on the rate to find the order for that reactant.

Step 1: Determine the order with respect to NH\textsubscript{4}\textsuperscript{+}. Compare Expt 1 and Expt 2 where \([NO_2^-]\) is constant at 0.020 M. Concentration of \(NH_4^+\) increases from 0.010 to 0.015 (1.5 times). Rate increases from 0.020 to 0.030 (1.5 times). Since \((1.5)^x = 1.5\), the order \(x = 1\).

Step 2: Determine the order with respect to NO\textsubscript{2}\textsuperscript{--}. Compare Expt 1 and Expt 3 where \([NH_4^+]\) is constant at 0.010 M. Concentration of \(NO_2^-\) decreases from 0.020 to 0.010 (halved). Rate decreases from 0.020 to 0.005 (\(1/4\)th). Since \((1/2)^y = 1/4\), then \((1/2)^y = (1/2)^2\), so order \(y = 2\).

Step 3: Construct the final rate law. Combining both orders: Rate = \(k [NH_4^+]^1 [NO_2^-]^2\). This matches Option (C).