Consider the following reaction between mercury(II) chloride and oxalate ion: $$ 2 \mathrm{HgCl}_{2}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \longrightarrow 2 \mathrm{Cl}^{-}(a q)+2 \mathrm{CO}_{2}(g)+\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s) $$ The initial rate of this reaction was determined for several concentrations of \(\mathrm{HgCl}_{2}\) and \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\), and the following rate data were obtained for the rate of disappearance of \(\mathrm{C}_{2} \mathrm{O}_{4}{ }^{2-}\) : $$ \begin{array}{llll} \hline \text { Experiment } & {\left[\mathrm{HgCl}_{2}\right](M)} & {\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right](M)} & \text { Rate }(M / \mathrm{s}) \\ \hline 1 & 0.164 & 0.15 & 3.2 \times 10^{-5} \\ 2 & 0.164 & 0.45 & 2.9 \times 10^{-4} \\ 3 & 0.082 & 0.45 & 1.4 \times 10^{-4} \\ 4 & 0.246 & 0.15 & 4.8 \times 10^{-5} \\ \hline \end{array} $$ (a) What is the rate law for this reaction? (b) What is the value of the rate constant with proper units? (c) What is the reaction rate when the initial concentration of \(\mathrm{HgCl}_{2}\) is \(0.100 \mathrm{M}\) and that of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\) is \(0.25 \mathrm{M}\) if the temperature is the same as that used to obtain the data shown?

Step by step solution

01

Determine the order with respect to each reactant

Let's start by comparing experiment 1 (E1) and experiment 2 (E2). The concentrations of HgCl2 remain constant at 0.164 M, while the concentration of \(C_2O_4^{2-}\) increases by a factor of 3 (0.15 M to 0.45 M). The rate increases from \(3.2 \times 10^{-5}\) M/s to \(2.9 \times 10^{-4}\) M/s, which is a factor of approximately 9. Similarly, we can compare experiment 1 (E1) and experiment 4 (E4). This time, the concentration of \(C_2O_4^{2-}\) remains constant at 0.15 M, while the concentration of HgCl2 increases by a factor of 1.5 (0.164 M to 0.246 M). The rate increases from \(3.2 \times 10^{-5}\) M/s to \(4.8 \times 10^{-5}\) M/s, which is a factor of 1.5. Now we can determine the order of the reaction for each reactant. Since the rate increases by a factor of 9 when the concentration of \(C_2O_4^{2-}\) is tripled, the order with respect to \(C_2O_4^{2-}\) must be 2. Likewise, since the rate increases by a factor of 1.5 when the concentration of HgCl2 increases by a factor of 1.5, the order with respect to HgCl2 must be 1.

02

Write the rate law

Now that we know the order of the reaction for each reactant, we can write the rate law: Rate = \(k[HgCl_2]^1[C_2O_4^{2-}]^2\)

03

Calculate the rate constant (k)

We can use one of the experiments to determine the rate constant. Let's use experiment 1 (E1) for this purpose. From E1, we have: Rate = \(3.2 \times 10^{-5}\) M/s, [HgCl2] = 0.164 M, [\(C_2O_4^{2-}\)] = 0.15 M Plug in these values into the rate law: \(3.2 \times 10^{-5}\) M/s = \(k(0.164 \thinspace M)^1(0.15 \thinspace M)^2\) Solve for k: k ≈ \(2.4 \times 10^{2}\) M⁻²s⁻¹

04

Calculate the reaction rate for given initial concentrations

To find the reaction rate when [HgCl2] = 0.100 M and [\(C_2O_4^{2-}\)] = 0.25 M, plug in these concentrations and the value of k into the rate law: Rate = \((2.4 \times 10^{2}\) M⁻²s⁻¹)(0.100 M)(0.25 M)²\) Rate = \(1.5 \times 10^{-4}\) M/s So, the answers are: (a) Rate law: Rate = \(k[HgCl_2]^1[C_2O_4^{2-}]^2\) (b) Rate constant, k ≈ \(2.4 \times 10^{2}\) M⁻²s⁻¹ (c) Reaction rate: \(1.5 \times 10^{-4}\) M/s

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