Chemistry, 16.12.2019 19:31 pearpeaerrr1993

Consider the following reaction between mercury(ii) chloride and oxalate ion: 2hgcl2(aq)+c2o2−4(aq)→2cl−(aq)+2co2 (g)+hg2cl2(s) the initial rate of this reaction was determined for several concentrations of hgcl2 and c2o2−4, and the following rate data were obtained for the rate of disappearance of c2o2−4:

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Answer from: hlgerardip4wbhx

The question is incomplete, here is a complete question.

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

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

The initial rate of this reaction was determined for several concentrations of HgCl_2 and $C_2O_2^{4-}$ , and the following rate data were obtained for the rate of disappearance of $C_2O_2^{4-}$ :

Experiment HgCl_2(M) $C_2O_2^{4-}(M)$ Rate (M/s)

1 0.164 0.15 3.2×10⁻⁵

2 0.164 0.45 2.9×10⁻⁴

3 0.082 0.45 1.4×10⁻⁴

4 0.246 0.15 4.8×10⁻⁵

What is the reaction rate when the concentration of HgCl_2 is 0.135 M and that of $C_2O_2^{4-}$ is 0.40 M , if the temperature is the same as that used to obtain the data shown?

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, $1.9\times 10^{-4}M/s$

Answer from: Quest

The breaking apart of glucose, which involves spontaneous reactions

Answer from: Quest

Sink to the bottom. addition of potassium nitrate past it’s dissolution capacity at a particular temperature will cause the extra to just sink to the bottom of the glass.however, if you were to raise the temperature of the solution prior to dissolving it, and then slowly decrease the temperature after the dissolution, you would create a “supersaturated” solution - a solution that has more solute dissolved into it than should be allowed. if you add any extra solute past this, the entire solution would “crystallize,” and this is how you get those cool videos on youtube.