Express the rate of reaction in terms of the change in concentration of each of the reactants and products: A(g) + 2B(g) → C(g) Rate = − Δ[A] Δt = − 2 1 Δ[B] Δt = Δ[C] Δt Rate = − Δ[A] Δt = − Δ[B] Δt = Δ[C] Δt Rate = − Δ[A] Δt = − 1 2 Δ[B] Δt = Δ[C] Δt When [B] is decreasing at 0.28 mol/L·s, how fast is [A] decreasing?

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Alleei Alleei · Answer 1 · 2019-11-11T08:12:15+01:00

Answer :

The correct expression for rate of reaction is:

[tex]Rate=-\frac{d[A]}{dt}=-\frac{1}{2}\frac{d[B]}{dt}=+\frac{d[C]}{dt}[/tex]

The [A] decreasing at 0.14 mol/L.s

Explanation :

The given rate of reaction is,

[tex]A(g)+2B(g)\rightarrow C(g)[/tex]

The expression for rate of reaction :

[tex]\text{Rate of disappearance of A}=-\frac{d[A]}{dt}[/tex]

[tex]\text{Rate of disappearance of B}=-\frac{1}{2}\frac{d[B]}{dt}[/tex]

[tex]\text{Rate of formation of C}=+\frac{d[C]}{dt}[/tex]

So,

[tex]Rate=-\frac{d[A]}{dt}=-\frac{1}{2}\frac{d[B]}{dt}=+\frac{d[C]}{dt}[/tex]

Now we have to calculate how fast is [A] decreasing.

Given:

[tex]-\frac{d[B]}{dt}=0.28mol/L.s[/tex]

[tex]-\frac{1}{2}\frac{d[B]}{dt}=-\frac{1}{2}\times 0.28mol/L.s=-0.14mol/L.s[/tex]

[tex]-\frac{d[A]}{dt}=-\frac{1}{2}\frac{d[B]}{dt}=-0.14mol/L.s[/tex]

Hence, the [A] decreasing at 0.14 mol/L.s