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Sagot :
We'll work through the following steps to determine the rate law and the rate constant, given provided reaction data.
### Part A: Determining the Rate Law
The general rate law for the reaction can be expressed as:
[tex]\[ \text{rate} = k [\text{Hg}_2\text{Cl}_2]^{m} [\text{C}_2\text{O}_4^{2-}]^{n} \][/tex]
To determine the orders of reaction, [tex]\( m \)[/tex] and [tex]\( n \)[/tex], we compare rates from different experiments where concentrations of one reactant change while the other remains constant.
We use logarithms to transform the rate equation into a linear form to make it easier to solve for [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
[tex]\[ \ln(\text{rate}) = \ln(k) + m\ln([\text{Hg}_2\text{Cl}_2]) + n\ln([\text{C}_2\text{O}_4^{2-}]) \][/tex]
By performing linear regression analysis on the transformed data, we extract [tex]\( m \)[/tex] and [tex]\( n \)[/tex]. According to the calculations, the results are:
[tex]\[ m = -1.05 \][/tex]
[tex]\[ n = -0.30 \][/tex]
Therefore, the rate law is:
[tex]\[ \text{rate} = k [\text{Hg}_2\text{Cl}_2]^{-1.05} [\text{C}_2\text{O}_4^{2-}]^{-0.30} \][/tex]
### Part B: Determining the Rate Constant [tex]\( k \)[/tex] and Its Units
Given [tex]\( m = -1.05 \)[/tex] and [tex]\( n = -0.30 \)[/tex], we use experimental data to find the rate constant [tex]\( k \)[/tex].
Using the provided experimental data for one of the experiments (e.g., Experiment 1):
[tex]\[ [\text{Hg}_2\text{Cl}_2] = 0.164\, \text{M} \][/tex]
[tex]\[ [\text{C}_2\text{O}_4^{2-}] = 8.15\, \text{M} \][/tex]
[tex]\[ \text{rate} = 3.2 \times 10^{-5} \, \text{M/s} \][/tex]
We substitute these values into our rate equation to solve for [tex]\( k \)[/tex]:
[tex]\[ 3.2 \times 10^{-5} = k (0.164)^{-1.05} (8.15)^{-0.30} \][/tex]
From the calculations performed (see given results), we get:
[tex]\[ k = 8.86 \times 10^{-6} \, \text{M}^{1.35} \text{s}^{-1} \approx 8.9 \times 10^{-6} \, \text{M}^{1.35} \text{s}^{-1} \][/tex]
The units of [tex]\( k \)[/tex] are determined to maintain the rate constant consistent with the rate law:
Since [tex]\( m + n = -1.05 + (-0.30) = -1.35 \)[/tex], the rate constant [tex]\( k \)[/tex] unit becomes:
[tex]\[ k \text{ unit} = \frac{\text{M/s}}{ \text{M}^{-1.35} } = \text{M}^{1.35} \text{s}^{-1} \][/tex]
Summary:
- Part A: Rate Law:
[tex]\[ \text{rate} = k [\text{Hg}_2\text{Cl}_2]^{-1.05} [\text{C}_2\text{O}_4^{2-}]^{-0.30} \][/tex]
- Part B: Rate Constant [tex]\( k \)[/tex]:
[tex]\[ k = 8.9 \times 10^{-6} \, \text{M}^{1.35} \text{s}^{-1}, \][/tex]
These calculations provide a comprehensive determination of the reaction kinetics based on the experimental data.
### Part A: Determining the Rate Law
The general rate law for the reaction can be expressed as:
[tex]\[ \text{rate} = k [\text{Hg}_2\text{Cl}_2]^{m} [\text{C}_2\text{O}_4^{2-}]^{n} \][/tex]
To determine the orders of reaction, [tex]\( m \)[/tex] and [tex]\( n \)[/tex], we compare rates from different experiments where concentrations of one reactant change while the other remains constant.
We use logarithms to transform the rate equation into a linear form to make it easier to solve for [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
[tex]\[ \ln(\text{rate}) = \ln(k) + m\ln([\text{Hg}_2\text{Cl}_2]) + n\ln([\text{C}_2\text{O}_4^{2-}]) \][/tex]
By performing linear regression analysis on the transformed data, we extract [tex]\( m \)[/tex] and [tex]\( n \)[/tex]. According to the calculations, the results are:
[tex]\[ m = -1.05 \][/tex]
[tex]\[ n = -0.30 \][/tex]
Therefore, the rate law is:
[tex]\[ \text{rate} = k [\text{Hg}_2\text{Cl}_2]^{-1.05} [\text{C}_2\text{O}_4^{2-}]^{-0.30} \][/tex]
### Part B: Determining the Rate Constant [tex]\( k \)[/tex] and Its Units
Given [tex]\( m = -1.05 \)[/tex] and [tex]\( n = -0.30 \)[/tex], we use experimental data to find the rate constant [tex]\( k \)[/tex].
Using the provided experimental data for one of the experiments (e.g., Experiment 1):
[tex]\[ [\text{Hg}_2\text{Cl}_2] = 0.164\, \text{M} \][/tex]
[tex]\[ [\text{C}_2\text{O}_4^{2-}] = 8.15\, \text{M} \][/tex]
[tex]\[ \text{rate} = 3.2 \times 10^{-5} \, \text{M/s} \][/tex]
We substitute these values into our rate equation to solve for [tex]\( k \)[/tex]:
[tex]\[ 3.2 \times 10^{-5} = k (0.164)^{-1.05} (8.15)^{-0.30} \][/tex]
From the calculations performed (see given results), we get:
[tex]\[ k = 8.86 \times 10^{-6} \, \text{M}^{1.35} \text{s}^{-1} \approx 8.9 \times 10^{-6} \, \text{M}^{1.35} \text{s}^{-1} \][/tex]
The units of [tex]\( k \)[/tex] are determined to maintain the rate constant consistent with the rate law:
Since [tex]\( m + n = -1.05 + (-0.30) = -1.35 \)[/tex], the rate constant [tex]\( k \)[/tex] unit becomes:
[tex]\[ k \text{ unit} = \frac{\text{M/s}}{ \text{M}^{-1.35} } = \text{M}^{1.35} \text{s}^{-1} \][/tex]
Summary:
- Part A: Rate Law:
[tex]\[ \text{rate} = k [\text{Hg}_2\text{Cl}_2]^{-1.05} [\text{C}_2\text{O}_4^{2-}]^{-0.30} \][/tex]
- Part B: Rate Constant [tex]\( k \)[/tex]:
[tex]\[ k = 8.9 \times 10^{-6} \, \text{M}^{1.35} \text{s}^{-1}, \][/tex]
These calculations provide a comprehensive determination of the reaction kinetics based on the experimental data.
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