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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

Respuesta :

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Answer:

8.88 x 10⁻² M/s

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