A second reaction mixture was made up in the following way: (please explain how)
ID: 475354 • Letter: A
Question
A second reaction mixture was made up in the following way: (please explain how)
20 mL of 4.0M acetone + 10 mL of 1.0 M HCl + 10 mL 0f 0.0050M I2 + 10 mL H2O
a. What is the initial concentration of acetone, H+ ion, I2 in the reaction mixture?
b. It took 120 seconds for the I2 color to disappear from the reaction mixture when it occurred at the same temperature as the reaction in Problem2. What was the rate of the reaction?
Write equation 3 as it would apply to the second reaction mixture:
rate =
Divide the equation in Part b by the equation in Problem 2b. The resulting equation should have the ratio of the two rates on the left side and a ratio of acetone concentration raised to the m power on the right. Write the resulting equation and solve for the value of m, the order of the reaction with respect to acetone. (Round off the value of m to the nearest integer.)
Explanation / Answer
Answer (a)
First, lets find the initial concentration of Acetone
20 ml * (1ml/1000L) * ( 4 mol/1Liter) * ( 1 / 0.05 L) = 1.6 M
(20 ml) * (coverting ml to Liter) * (molarity of the acetone) * ( 1 / volume of mixture) = Mol acetone/ L mixture
Second then find the initial concentration of H+ ion
10 ml * (1ml/1000L) * ( 1 mol/1Liter) * ( 1 / 0.05 L) = 0.2 M
lets find the initial concentration of I2.
10 ml * (1ml/1000L) * (0.005 mol/1Liter) * ( 1 / 0.05 L) = 0.001 M
Answer(b)
The I2 color disappears when all the I2 is used up, so when concentration = 0
0 - 0.0005 = - 0.0005 = 0.0005 loss in M for I2 color to dissapear, this is where the reaction rate comes in.
Here It took 120 seconds for the I2 color to disappear from the reaction mixture when it occurred at the same temperature as the reaction so the rate of reaction will be:
0.0005 M / 120 seconds = 4.17 * 10-6 M / second
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