H2PO4-(aq)+OH-(aq) -> HPO42-(aq)+H2O
Experiment/[H2PO4-](M)/[OH-](M)/ Initial Rate(mol/L·min)
1 0.0031 0.00041 0.0021
2 0.0031 0.00082 0.0084
3 0.0093 0.00041 0.0063
4 ? 0.00034 0.0021
(second rows are moles of H2PO4, third rows are moles of OH, and the last rows are initial rate).
I am supposed to find the rate law, but I am having trouble. When I was trying to find x and y, the numbers were so bizarre. For x, it was 0.75 = 3^x and for y, it was 4 = 20^y. Did I do something wrong in the process? help me!
You must have. I ran through y (the exponent for OH^- and arrived at a value of 2. For x, the exponent of H2PO4^- I found 1.
For y, divide equation 2 by 1.
For x, divide equation 3 by 1.
If you can't find the error, post your work and I'll find it for you.
To determine the rate law for this reaction, we need to use the experimental data provided. The rate law is an equation that relates the initial rate of the reaction to the concentrations of the reactants. In this case, we need to determine the exponents (x and y) for the reactants [H2PO4-] and [OH-] in the rate law equation.
Let's analyze the data to determine the rate law:
Experiment / [H2PO4-] (M) / [OH-] (M) / Initial Rate (mol/L·min)
1 / 0.0031 / 0.00041 / 0.0021
2 / 0.0031 / 0.00082 / 0.0084
3 / 0.0093 / 0.00041 / 0.0063
4 / ? / 0.00034 / 0.0021
From the given data, we can see that experiment 4 was conducted at the same concentration of OH- as experiment 1, but with an unknown concentration of H2PO4-. We can use the data from experiments 1-3 to determine the exponents x and y.
Using the method of initial rates, compare experiments 1 and 2:
(rate_exp2)/(rate_exp1) = [(0.0084 M/min)/(0.0021 M/min)] = (0.0031 M/0.0031 M)^x * (0.00082 M/0.00041 M)^y
Simplifying the equation, we get:
4 = 2^x * 2^y
Since 4 = 2^2, we can infer that x + y = 2.
Similarly, compare experiments 1 and 3:
(rate_exp3)/(rate_exp1) = [(0.0063 M/min)/(0.0021 M/min)] = (0.0093 M/0.0031 M)^x * (0.00041 M/0.00041 M)^y
Simplifying the equation, we get:
3 = 3^x * 1^y
Since 3 = 3^1, we can infer that x = 1 and y = 0.
Now, we have determined the values of x and y for the rate law equation:
rate = k[H2PO4-]^1[OH-]^0
Simplifying further, we get:
rate = k[H2PO4-]
Therefore, the rate law for this reaction is: rate = k[H2PO4-].
Note: The value of y is 0 because [OH-] does not affect the rate of the reaction.