Question

Given the following balanced equation, determine the rate of reaction with respect to [Cl2]. If the rate of Cl2 loss is 4.44 × 10-2 M/s, what is the rate of formation of NOCl?2 NO(g) + Cl2(g) → 2 NOCl(g)1.11 × 10-1 M/s2.22 × 10-2 M/s8.88 × 10-2 M/s1.61 × 10-2 M/s4.44 × 10-2 M/s

Answer 1

8.88 x 10⁻² M/s

Step-by-step explanation:

The rate of reaction for:

NO(g) + Cl₂ (g) ⇒ 2NOCl(g)

is rate = -ΔNO/Δt = -ΔCl2/Δt = 1/2 ΔNOCl/Δt

so ΔNOCl/Δt = 2 ΔCl2/Δt = 2 x 4.44 × 10⁻² M/s = 8.88 x 10⁻² M/s

In general given a reaction

aA + bB ⇒ cC + dD

rate = -1/a ΔA/Δt = -1/b ΔB/Δt = 1/c ΔC/Δt = 1/d ΔD/Δt