Rate = k’[CV], where k’ = k[NaOH]y
Look at the graphs for Run 1 and Run 2 that pertain to the order of the reaction with respect to [CV]. The slope of the graphs is equal to k’, which is itself equal to k[NaOH]^y. In the first run, [NaOH] = 0.010 M after dilution, and in the second run, [NaOH] = 0.020 M after dilution. Use this information
to find the value for “y”.
I have the graphs but the wording for this problem on setting it up and finding initial rate is confusing.
To find the value of "y" in the rate equation, you need to compare the slopes of two graphs representing the rates of reaction at different concentrations of sodium hydroxide ([NaOH]).
Here's how you can set up and solve the problem:
Step 1: Understand the setup
You are given that the rate equation is Rate = k’[CV], where k’ = k[NaOH]^y. In this equation, "Rate" represents the rate of the reaction, "[CV]" represents the concentration of the reactant CV, "k'" is the rate constant, "k" is a proportionality constant, and "[NaOH]" represents the concentration of sodium hydroxide.
Step 2: Analyze the graphs
You have two graphs, one for Run 1 and one for Run 2. Plot the rate of the reaction on the y-axis and the concentration of CV on the x-axis.
Step 3: Determine the slopes
The slopes of the graphs represent k' (the rate constant multiplied by the concentration of sodium hydroxide raised to the power of "y") for each run. Label the slopes as m1 and m2 for Run 1 and Run 2, respectively.
Step 4: Set up the equations
From the given information, write the expressions for k' for each run:
k' = m1 = k[NaOH1]^y, where [NaOH1] = 0.010 M after dilution (Run 1)
k' = m2 = k[NaOH2]^y, where [NaOH2] = 0.020 M after dilution (Run 2)
Step 5: Solve for "y"
Divide the two equations to eliminate the rate constant "k":
m1/m2 = (k[NaOH1]^y) / (k[NaOH2]^y)
Since k is canceled out, you are left with:
m1/m2 = ([NaOH1]/[NaOH2])^y
Substitute the given values:
m1/m2 = (0.010 M / 0.020 M)^y
Step 6: Calculate "y"
To solve for "y", take the logarithm of both sides of the equation and rearrange:
log(m1/m2) = y * log([NaOH1]/[NaOH2])
Now you can solve for "y" using the formula:
y = log(m1/m2) / log([NaOH1]/[NaOH2])
Plug in the calculated values of m1, m2, [NaOH1], and [NaOH2] to find the value of "y" in the rate equation.