Consider the following single step reaction in gas phase at constant temperature.
2 A_((g))+B_((g)) → C_((g))
The initial rate of the reaction is recorded as r₁ when the reaction starts with 1.5 atm pressure of A and 0.7 atm pressure of B. After some time, the rate r₂ is recorded when the pressure of C becomes 0.5 atm. The ratio r₁: r₂ is _________ × 10⁻¹. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

A single-step reaction lets you read the rate law off the equation, so the rate is k p_A² p_B.

1. Rate law. A single-step (elementary) reaction has orders equal to its stoichiometric coefficients. So the rate is r = k p_A² p_B: second order in A and first order in B.
2. Pressures later on. Forming 0.5 atm of C uses 2 × 0.5 = 1.0 atm of A and 0.5 atm of B. That leaves p_A = 1.5 - 1.0 = 0.5 atm and p_B = 0.7 - 0.5 = 0.2 atm.

   r₁/r₂ = ((1.5)² × 0.7)/((0.5)² × 0.2) = 1.575/0.05 = 31.5

3. So. r₁ : r₂ = 31.5 = 315 × 10⁻¹, so the answer is 315.

The common slip: 105. Treating the reaction as first order in A gives (1.5 × 0.7)/(0.5 × 0.2) = 10.5, which you would report as 105. The coefficient 2 on A makes the rate depend on p_A².

The idea to keep. Single step means elementary, so the coefficients become the orders. Then use the stoichiometry to find how much of each reactant is used up before you put the pressures into the rate law.

Answer from the NTA final answer key. Solution by PrepWiser.