Question

Consider the following reaction between mercury(II) chloride and oxalate ion:

Answer 1

Answer : The reaction rate will be, $1.9\times 10^{-4}M/s$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$2HgCl_2(aq)+C_2O_2^{4-}(aq)\rightarrow 2Cl^-(aq)+2CO_2(g)+HgCl_2(s)$

Rate law expression for the reaction:

$\text{Rate}=k[HgCl_2]^a[C_2O_2^{4-}]^b$

where,

a = order with respect to $HgCl_2$

b = order with respect to $C_2O_2^{4-}$

Expression for rate law for first observation:

$3.2\times 10^{-5}=k(0.164)^a(0.15)^b$ ....(1)

Expression for rate law for second observation:

$2.9\times 10^{-4}=k(0.164)^a(0.45)^b$ ....(2)

Expression for rate law for third observation:

$1.4\times 10^{-4}=k(0.082)^a(0.45)^b$ ....(3)

Expression for rate law for fourth observation:

$4.8\times 10^{-5}=k(0.246)^a(0.15)^b$ ....(4)

Dividing 1 from 2, we get:

$\frac{2.9\times 10^{-4}}{3.2\times 10^{-5}}=\frac{k(0.164)^a(0.45)^b}{k(0.164)^a(0.15)^b}\\\\9=3^b\\(3)^2=3^b\\b=2$

Dividing 3 from 2, we get:

$\frac{2.9\times 10^{-4}}{1.4\times 10^{-4}}=\frac{k(0.164)^a(0.45)^b}{k(0.082)^a(0.45)^b}\\\\2=2^a\\a=1$

Thus, the rate law becomes:

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

Now, calculating the value of 'k' by using any expression.

Putting values in above rate law, we get:

$3.2\times 10^{-5}=k(0.164)^1(0.15)^2$

$k=8.7\times 10^{-3}M^{-2}s^{-1}$

Now we have to determine the reaction rate when the concentration of $HgCl_2$ is 0.135 M and that of $C_2O_2^{-4}$ is 0.40 M.

$\text{Rate}=k[HgCl_2]^1[C_2O_2^{4-}]^2$

$\text{Rate}=(8.7\times 10^{-3})\times (0.135)^1\times (0.40)^2$

$\text{Rate}=1.9\times 10^{-4}M/s$

Therefore, the reaction rate will be, $1.9\times 10^{-4}M/s$