Consider the reaction: 2A(g)+B(g)→3C(g).

Part A

Determine the expression for the rate of the reaction with respect to each of the reactants and products.
a) Rate=−13Δ[A]Δt=−Δ[B]Δt=12Δ[C]Δt
b) Rate=−12Δ[A]Δt=−Δ[B]Δt=13Δ[C]Δt
c) Rate=−Δ[A]Δt=−12Δ[B]Δt=13Δ[C]Δt
d) Rate=12Δ[A]Δt=12Δ[B]Δt=13Δ[C]Δt
Part B

When A is decreasing at a rate of 0.100 M⋅s−1 , how fast is B decreasing?
Part C

How fast is C increasing?

Question

contestada

Respuesta :

tirana tirana · Answer 1 · 2019-09-16T07:13:57+02:00

Answer:

Part A

[tex]Rate = -\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t} = \frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]

Part B

[tex]-\frac{\Delta B}{\Delta t}= 0.0500 M s^{-1} [/tex]

Part C

[tex]\frac{\Delta C}{\Delta t} = 0.15 M s^{-1}[/tex]

Explanation:

For a general reaction,

[tex]aA(g) + bB(g) \rightarrow cC(g)[/tex]

Rate is given by:

Rate: [tex]Rate = -\frac{1}{a}\frac{\Delta A}{\Delta t} =-\frac{1}{b}\frac{\Delta B}{\Delta t} = \frac{1}{c}\frac{\Delta C}{\Delta t}[/tex]

So, for the given reaction:

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

[tex]Rate = -\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t} = \frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]

Part B

[tex]-\frac{1}{2}\frac{\Delta A}{\Delta t} =-\frac{\Delta B}{\Delta t}[/tex]

Given: [tex]-\frac{\Delta A}{\Delta t} = 0.100\ Ms^{-1}[/tex]

[tex] \frac{1}{2}\frac{0.100}{\Delta t} =-\frac{\Delta B}{\Delta t}[/tex]

[tex]-\frac{\Delta B}{\Delta t}[/tex] = 0.0500 M s^-1

Part C

[tex]-\frac{\Delta B}{\Delta t} =\frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]

[tex]0.0500 = \frac{1}{3}\frac{\Delta C}{\Delta t}[/tex]

[tex]\frac{\Delta C}{\Delta t} = 3 \times 0.0500 = 0.15 M s^{-1}[/tex]