Question 59e58

Mar 15, 2015

$2 C r {O}_{4 \left(a q\right)}^{2 -} + 2 {H}_{\left(a q\right)}^{2 +} \to C r {O}_{7 \left(a q\right)}^{2 -} + {H}_{2} {O}_{\left(l\right)}$

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 $C r {O}_{7}^{2 -}$ must be equal to the decrease in concentration of $C r {O}_{4}^{2 -}$ 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 $a A \to b B$ (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 $B$, $r a t {e}_{B}$, is

$A = - \frac{a}{b} \cdot {\text{rate}}_{B}$

In your case, the rate of formation of $C r {O}_{7}^{2 -}$ is ${\text{0.14 mol" * "L"^(-1) * "s}}^{- 1}$, which means that the rate of disappearance of $C r {O}_{4}^{2 -}$ will be

${\text{rate"_("chromate") = -2/1 * "0.14 mol" * "L"^(-1) * "s}}^{- 1}$

${\text{rate"_("chromate") = - "0.28 mol" * "L"^(-1) * "s}}^{- 1}$

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

The rate of reaction in this case will be

$r a t {e}_{\text{rxn}} = - r a t {e}_{C r {O}_{4}^{2 -}} = r a t {e}_{C r {O}_{7}^{2 -}}$

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