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To determine the rate law for the given reaction, we need to identify the reaction orders with respect to each reactant, [tex]\( \text{[HgCl}_2\text{]} \)[/tex] and [tex]\( \text{[C}_2\text{O}_4^{2-}\text{]} \)[/tex]. We'll use the method of initial rates with the provided experimental data:
| Experiment | [tex]\( \text{HgCl}_2 \, (\text{M}) \)[/tex] | [tex]\( \text{C}_2\text{O}_4^{2-} \, (\text{M}) \)[/tex] | Rate (M/s) |
|------------|-----------------|------------------|--------------------------|
| 1 | 0.764 | 0.15 | [tex]\( 3.2 \times 10^{-4} \)[/tex] |
| 2 | 0.164 | 0.45 | [tex]\( 2.9 \times 10^{-4} \)[/tex] |
| 3 | 0.062 | 0.45 | [tex]\( 1.4 \times 10^{-4} \)[/tex] |
| 4 | 0.246 | 0.15 | [tex]\( 4.8 \times 10^{-1} \)[/tex] |
Assume the rate law is of the form:
[tex]\[ \text{Rate} = k \left[\text{HgCl}_2\right]^m \left[\text{C}_2\text{O}_4^{2-}\right]^n \][/tex]
### Step 1: Determine the order with respect to [tex]\( \text{HgCl}_2 \)[/tex] (m)
Compare experiments 1 and 3, where [tex]\( \text{[C}_2\text{O}_4^{2-}\text{]} \)[/tex] is constant and [tex]\( \text{[HgCl}_2\text{]} \)[/tex] changes.
[tex]\[ \frac{\text{Rate}_1}{\text{Rate}_3} = \frac{k \left[ \text{HgCl}_2 \right]_1^m \left[ \text{C}_2 \text{O}_4^{2-} \right]_1^n}{k \left[ \text{HgCl}_2 \right]_3^m \left[ \text{C}_2 \text{O}_4^{2-} \right]_3^n} \][/tex]
Since [tex]\( \left[ \text{C}_2 \text{O}_4^{2-} \right] \)[/tex] is constant, it cancels out:
[tex]\[ \frac{\text{Rate}_1}{\text{Rate}_3} = \left( \frac{ \left[ \text{HgCl}_2 \right]_1}{\left[ \text{HgCl}_2 \right]_3} \right)^m \][/tex]
Substitute the values from experiments 1 and 3:
[tex]\[ \frac{3.2 \times 10^{-4}}{1.4 \times 10^{-4}} = \left( \frac{0.764}{0.062} \right)^m \][/tex]
[tex]\[ 2.2857 = 12.3226^m \][/tex]
Taking the logarithm of both sides:
[tex]\[ \log(2.2857) = m \log(12.3226) \][/tex]
[tex]\[ m \approx \frac{\log(2.2857)}{\log(12.3226)} \approx 0.3173 \approx 0.3 \][/tex]
### Step 2: Determine the order with respect to [tex]\( \text{C}_2\text{O}_4^{2-} \)[/tex] (n)
Compare experiments 2 and 3, where [tex]\( \text{[HgCl}_2\text{]} \)[/tex] is constant and [tex]\( \text{[C}_2\text{O}_4^{2-}\text{]} \)[/tex] changes.
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_3} = \frac{k \left[ \text{HgCl}_2 \right]_2^m \left[ \text{C}_2 \text{O}_4^{2-} \right]_2^n}{k \left[ \text{HgCl}_2 \right]_3^m \left[ \text{C}_2 \text{O}_4^{2-} \right]_3^n} \][/tex]
Since [tex]\( \left[ \text{HgCl}_2 \right] \)[/tex] is constant, it cancels out:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_3} = \left( \frac{\left[ \text{C}_2 \text{O}_4^{2-} \right]_2}{\left[ \text{C}_2 \text{O}_4^{2-} \right]_3} \right)^n \][/tex]
Substitute the values from experiments 2 and 3:
[tex]\[ \frac{2.9 \times 10^{-4}}{1.4 \times 10^{-4}} = \left(\frac{0.45}{0.45}\right)^n \][/tex]
The ratio of concentrations is 1, indicating [tex]\(n=0\)[/tex] if simplified but we must validate further considering experimental variations, hence we test between 1 and 2:
The magnitudes may verify [tex]\(n=1\)[/tex].
### Step 3: Determine the rate constant [tex]\(k\)[/tex]
Using experiment 1 with found orders [tex]\( m=0.3, n= 0.4\)[/tex]
[tex]\[ \text{Rate}_1 = k \left[\text{HgCl}_2 \right]_1^{0.3} \left[\text{C}_2\text{O}_4^{2-}\right]_1^{1} \][/tex]
[tex]\[ 3.2 \times 10^{-4} = k \left(0.764\right)^{0.3} \left(0.15\right) \][/tex]
Calculate [tex]\( k \)[/tex]:
\[
k= 3.2 \times 10^{-4} / 0.1189 = 2.691 \times 10^{-3}
Proper Units \(M /s means =2.7 \approx
Answer:
Exact Law even Close values suggests
Rate = k [HCNI2] 0.3 [C _2]
| Experiment | [tex]\( \text{HgCl}_2 \, (\text{M}) \)[/tex] | [tex]\( \text{C}_2\text{O}_4^{2-} \, (\text{M}) \)[/tex] | Rate (M/s) |
|------------|-----------------|------------------|--------------------------|
| 1 | 0.764 | 0.15 | [tex]\( 3.2 \times 10^{-4} \)[/tex] |
| 2 | 0.164 | 0.45 | [tex]\( 2.9 \times 10^{-4} \)[/tex] |
| 3 | 0.062 | 0.45 | [tex]\( 1.4 \times 10^{-4} \)[/tex] |
| 4 | 0.246 | 0.15 | [tex]\( 4.8 \times 10^{-1} \)[/tex] |
Assume the rate law is of the form:
[tex]\[ \text{Rate} = k \left[\text{HgCl}_2\right]^m \left[\text{C}_2\text{O}_4^{2-}\right]^n \][/tex]
### Step 1: Determine the order with respect to [tex]\( \text{HgCl}_2 \)[/tex] (m)
Compare experiments 1 and 3, where [tex]\( \text{[C}_2\text{O}_4^{2-}\text{]} \)[/tex] is constant and [tex]\( \text{[HgCl}_2\text{]} \)[/tex] changes.
[tex]\[ \frac{\text{Rate}_1}{\text{Rate}_3} = \frac{k \left[ \text{HgCl}_2 \right]_1^m \left[ \text{C}_2 \text{O}_4^{2-} \right]_1^n}{k \left[ \text{HgCl}_2 \right]_3^m \left[ \text{C}_2 \text{O}_4^{2-} \right]_3^n} \][/tex]
Since [tex]\( \left[ \text{C}_2 \text{O}_4^{2-} \right] \)[/tex] is constant, it cancels out:
[tex]\[ \frac{\text{Rate}_1}{\text{Rate}_3} = \left( \frac{ \left[ \text{HgCl}_2 \right]_1}{\left[ \text{HgCl}_2 \right]_3} \right)^m \][/tex]
Substitute the values from experiments 1 and 3:
[tex]\[ \frac{3.2 \times 10^{-4}}{1.4 \times 10^{-4}} = \left( \frac{0.764}{0.062} \right)^m \][/tex]
[tex]\[ 2.2857 = 12.3226^m \][/tex]
Taking the logarithm of both sides:
[tex]\[ \log(2.2857) = m \log(12.3226) \][/tex]
[tex]\[ m \approx \frac{\log(2.2857)}{\log(12.3226)} \approx 0.3173 \approx 0.3 \][/tex]
### Step 2: Determine the order with respect to [tex]\( \text{C}_2\text{O}_4^{2-} \)[/tex] (n)
Compare experiments 2 and 3, where [tex]\( \text{[HgCl}_2\text{]} \)[/tex] is constant and [tex]\( \text{[C}_2\text{O}_4^{2-}\text{]} \)[/tex] changes.
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_3} = \frac{k \left[ \text{HgCl}_2 \right]_2^m \left[ \text{C}_2 \text{O}_4^{2-} \right]_2^n}{k \left[ \text{HgCl}_2 \right]_3^m \left[ \text{C}_2 \text{O}_4^{2-} \right]_3^n} \][/tex]
Since [tex]\( \left[ \text{HgCl}_2 \right] \)[/tex] is constant, it cancels out:
[tex]\[ \frac{\text{Rate}_2}{\text{Rate}_3} = \left( \frac{\left[ \text{C}_2 \text{O}_4^{2-} \right]_2}{\left[ \text{C}_2 \text{O}_4^{2-} \right]_3} \right)^n \][/tex]
Substitute the values from experiments 2 and 3:
[tex]\[ \frac{2.9 \times 10^{-4}}{1.4 \times 10^{-4}} = \left(\frac{0.45}{0.45}\right)^n \][/tex]
The ratio of concentrations is 1, indicating [tex]\(n=0\)[/tex] if simplified but we must validate further considering experimental variations, hence we test between 1 and 2:
The magnitudes may verify [tex]\(n=1\)[/tex].
### Step 3: Determine the rate constant [tex]\(k\)[/tex]
Using experiment 1 with found orders [tex]\( m=0.3, n= 0.4\)[/tex]
[tex]\[ \text{Rate}_1 = k \left[\text{HgCl}_2 \right]_1^{0.3} \left[\text{C}_2\text{O}_4^{2-}\right]_1^{1} \][/tex]
[tex]\[ 3.2 \times 10^{-4} = k \left(0.764\right)^{0.3} \left(0.15\right) \][/tex]
Calculate [tex]\( k \)[/tex]:
\[
k= 3.2 \times 10^{-4} / 0.1189 = 2.691 \times 10^{-3}
Proper Units \(M /s means =2.7 \approx
Answer:
Exact Law even Close values suggests
Rate = k [HCNI2] 0.3 [C _2]
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