Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

a. rate=12Δ[HBr]Δt=−Δ[H2]Δt=−Δ[Br2]Δtrate=12Δ[HBr]Δt=−Δ[H2]Δt=−Δ[Br2]Δt
b. rate=−Δ[HBr]Δt=12Δ[H2]Δt=12Δ[Br2]Δtrate=−Δ[HBr]Δt=12Δ[H2]Δt=12Δ[Br2]Δt
c. rate=−12Δ[HBr]Δt=Δ[H2]Δt=Δ[Br2]Δtrate=−12Δ[HBr]Δt=Δ[H2]Δt=Δ[Br2]Δt
d. rate=Δ[HBr]Δt=−12Δ[H2]Δt=−12Δ[Br2]Δt

Respuesta :

Answer:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

[tex]2HBr(g)\rightarrow H_2(g)+Br_2(g)[/tex]

Thus, the rate is given as:

[tex]rate=-\frac{1}{2} \frac{\Delta [HBr]}{\Delta t}=\frac{\Delta [Br_2]}{\Delta t} =\frac{\Delta [H_2]}{\Delta t}[/tex]

It is necessary to remember that each concentration to time interval is divided into the stoichiometric coefficient, that is why HBr has a 1/2. Moreover, the concentration HBr is negative since it is a reactant and it has a negative rate due to its consumption.

Therefore, the answer is:

c. rate=−1/2Δ[HBr]/Δt=Δ[H2]/Δt=Δ[Br2]/Δt

Best regards.

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