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Given the following reaction: 2D(g) + 3E(g) + F(g) \longrightarrow⟶ 2G(g) + H(g) When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing? When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing? What is the rate of reaction?

Respuesta :

Answer:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

E decreseas 3/2 as fast as G increases = 0.30 M/s

Explanation:

Rate of reaction = -d[D] / 2dt  = -d[E]/ 3dt = -d[F]/dt  = d[G]/2dt = d[H]/dt

When the concentration of D is decreasing by 0.10 M/s, how fast is the concentration of H increasing:

Given data = d[D]/dt = 0.10 M/s

-d[D] / 2dt  = d[H]/dt

d[H]/dt = 0.05 M/s

The concentration of H is increasing, half as fast as D decreases: 0.05 mol L–1.s–1

When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing:

d[G] / 2dt  = -d[H]/3dt

E decreseas 3/2 as fast as G increases = 0.30 M/s

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