Question

Question #59e58

Answer 1

Start with the balanced chemical equation

`2CrO_(4(aq))^(2-) + 2H_((aq))^(2+) -> CrO_(7(aq))^(2-) + H_2O_((l))`

The rate of formation of a product, in your case of the dichromate ion, represents the amount of that compound formed per unit of time.

It is important to realize that the rate of formation for a product must be equal to the rate of disappearance of a reactant. In other words, the increase in the concentration of `CrO_7^(2-)` must be equal to the decrease in concentration of `CrO_4^(2-)` in the same period of time.

So, always keep an eye out for the stoichiometric coefficients for a given reaction, because the rate of reaction depends on stoichiometry.

If you look at a general form reaction `aA -> bB` (I'll use one reactant and one product because that is what you have to work with), the rate of disappearance of `A` when you know the rate of formation of `B`, `rate_B`, is

`A = -a/b * "rate"_B`

In your case, the rate of formation of `CrO_7^(2-)` is `"0.14 mol" * "L"^(-1) * "s"^(-1)`, which means that the rate of disappearance of `CrO_4^(2-)` will be

`"rate"_("chromate") = -2/1 * "0.14 mol" * "L"^(-1) * "s"^(-1)`

`"rate"_("chromate") = - "0.28 mol" * "L"^(-1) * "s"^(-1)`

The minus sign is used to symbolize the fact that this compound is being consumed.

The rate of reaction in this case will be

`rate_("rxn") = -rate_(CrO_4^(2-)) = rate_(CrO_7^(2-))`

`"rate"_("rxn") = -1/2 * (Delta[CrO_4^(2-)])/(Deltat) = 1/1 * (Delta[CrO_7^(2-)])/(Deltat)`