Given the following reaction: 3D(g) + E(g) + 2F(g) → 5G(g) + 4H(g)

(a) When the concentration of D is decreasing by 0.12 M/s, how fast is the concentration of H increasing?
(b) When the concentration of G is increasing by 0.20 M/s, how fast is the concentration of E decreasing?
(c) What is the rate of the reaction?

[tex]\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}[/tex]

Now, we answer:

(a)

[tex]\frac{d[H]}{dt}=4*\frac{1}{-3} *(-0.12M/s)=0.16M/s[/tex]

(b)

[tex]\frac{d[E]}{dt}=-1*\frac{1}{5} *(0.2M/s)=-0.04M/s[/tex]

(c) Since no initial data is specified, we could establish the rate of the reaction as based of the law of mass action:

[tex]r=\frac{1}{-3} \frac{d[D]}{dt} =\frac{1}{-1} \frac{d[E]}{dt} =\frac{1}{-2} \frac{d[F]}{dt} =\frac{1}{5} \frac{d[G]}{dt} =\frac{1}{4} \frac{d[H]}{dt}[/tex]

Thus, any of the available expressions are suitable to quantify the rate of the reaction.

Best regards.

aliasger2709 aliasger2709 · Answer 2 · 2022-02-26T05:59:33+01:00

The speed of the product formation is called the rate of the reaction. The concentration of H increases by 0.16 M/s, the concentration of E decreases by -0.04 M/s.

What is the rate of reaction?

The rate of the reaction is the speed or the rate at which the reaction occurs and the product is formed. It relates the time with the concentration of the reactants.

The required amount of the reactants in the reaction can be given by the law of mass action as,

[tex]\rm \dfrac{1}{-3} \dfrac{d[D]}{dt} = \rm \dfrac{1}{-1} \dfrac{d[E]}{dt} = \rm \dfrac{1}{-2} \dfrac{d[F]}{dt} = \rm \dfrac{1}{5} \dfrac{d[G]}{dt} = \rm \dfrac{1}{4} \dfrac{d[H]}{dt}[/tex]

The concentration rate of the H can be calculated as:

[tex]\begin{aligned}\rm \dfrac{d[H]}{dt} &= 4 \times \dfrac{1}{-3} \times (-0.12)\\\\&= 0.16\;\rm M/s\end{aligned}[/tex]

The concentration rate of the E can be calculated as:

[tex]\begin{aligned}\rm \dfrac{d[E]}{dt} &= -1 \times \dfrac{1}{5} \times (0.2)\\\\&= -0.04\;\rm M/s\end{aligned}[/tex]

The rate of the reaction is based on the law of mass action as no data of the initial concentration are given,

[tex]\rm r = \rm \dfrac{1}{-3} \dfrac{d[D]}{dt} = \rm \dfrac{1}{-1} \dfrac{d[E]}{dt} = \rm \dfrac{1}{-2} \dfrac{d[F]}{dt} = \rm \dfrac{1}{5} \dfrac{d[G]}{dt} = \rm \dfrac{1}{4} \dfrac{d[H]}{dt}[/tex]

Therefore, the concentration of H increased by 0.16 M/s and E decreased by -0.04 M/s.

Learn more about the rate of reaction here:

https://brainly.com/question/8437557