Consider the following data for the given reaction
2 HI_((g)) → H_(2( g))+I_(2( g))
The order of the reaction is _________.

Solution

Chemistry ma’am · 4 steps

Watch what the rate does each time [HI] doubles: that factor tells you the order.

1. Rate law. Write the rate as r = k[HI]ⁿ. Dividing two runs cancels k, so only the order n is left to find.

   r₂/r₁ = ([HI]₂/[HI]₁)ⁿ

2. Runs 1 and 2. [HI] goes from 0.005 to 0.01 mol L⁻¹, a factor of 2. The rate goes from 7.5 × 10⁻⁴ to 3.0 × 10⁻³ mol L⁻¹ s⁻¹, a factor of 4. So 2ⁿ = 4, giving n = 2.
3. Check runs 2, 3. [HI] doubles again, from 0.01 to 0.02 mol L⁻¹, and the rate goes from 3.0 × 10⁻³ to 1.2 × 10⁻² mol L⁻¹ s⁻¹, again a factor of 4. This confirms n = 2.
4. So. The rate law is r = k[HI]², and the order of the reaction is 2.

The common slip: 4. Taking the rate ratio itself as the order gives 4. The order is the power n in 2ⁿ = 4, which is 2.

The idea to keep. Order comes from experiment, not from the coefficients in the equation. If doubling a concentration multiplies the rate by 2ⁿ, the order in that reactant is n.

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