I'm really stuck with this one. Here's the rate law and the data that is needed. I appreciate any help
Rate= k[I2]X{[H+]Y[CH3COCH3]Z
Column 1 - Trial
Column 2 - Volume of 0.0010 M I2
Column 3 - Volume of 0.050 M HCl
Column 4 - Volume of 1.0 M acetone
Column 5 - Volume of water Temperature (degrees C)
Column 6 - Reaction time
A 5.0 10.0 10.0 25.0 25.0 130
B 10.0 10.0 10.0 20.0 25.0 249
C 10.0 20.0 10.0 10.0 25.0 128
D 10.0 10.0 20.0 10.0 25.0 131
E 10.0 10.0 10.0 20.0 42.4 38
What is the rate constant at 25.0 degrees C based on the data collected for trial B? Assume that the relative rate is equal to the actual rate in molar per second.
The 7th is reaction time in seconds.
To determine the rate constant at 25.0 degrees C based on the data collected for trial B, we first need to determine the order of reaction with respect to each reactant (X, Y, and Z) by comparing different trials. Here's the step-by-step process to find the rate constant:
Step 1: Compare the initial concentrations of I2, HCl, and acetone between trial B and another trial where only one reactant's concentration is changed. Let's compare trial B with trial A.
Comparing trial B and trial A:
- I2 concentration: Volume of 0.0010 M is the same in both trials.
- HCl concentration: Volume of 0.050 M is the same in both trials.
- Acetone concentration: Volume of 1.0 M is the same in both trials.
Since the volumes (concentrations) of I2, HCl, and acetone are the same in trial B and trial A, we can conclude that the order of the reaction with respect to each reactant is 1 (X = 1, Y = 1, Z = 1).
Step 2: Use the rate equation and the order of each reactant to calculate the rate constant (k).
Rate = k[I2]^X[H+]^Y[CH3COCH3]^Z
From the given data for trial B:
Rate (B) = k[I2]^1[H+]^1[CH3COCH3]^1
By substituting the given values for trial B into the rate equation, we get:
249 (B) = k(10.0)^1(10.0)^1(10.0)^1
Step 3: Solve for the rate constant (k).
249 = k(10.0)(10.0)(10.0)
k = 249 / (10.0)(10.0)(10.0)
k ≈ 0.249
Therefore, based on the data collected for trial B, the rate constant at 25.0 degrees C is approximately 0.249 (units depend on the units of the rate equation).
To find the rate constant at 25.0 degrees C based on the data collected for trial B, we need to first identify the values of X, Y, and Z in the rate law equation: Rate = k[I2]^X{[H+]^Y[CH3COCH3]^Z}.
From the data for trial B, we can identify the values:
Column 1 - Trial: B
Column 2 - Volume of 0.0010 M I2: 10.0
Column 3 - Volume of 0.050 M HCl: 10.0
Column 4 - Volume of 1.0 M acetone: 10.0
Column 5 - Volume of water Temperature (degrees C): 25.0
Column 6 - Reaction time: 249
Now, we can make a ratio of the rates of trial B and a reference trial where the concentrations are known:
(rate of B) / (rate of reference trial) = (k[I2]^X{[H+]^Y[CH3COCH3]^Z}) / (k[I2]_ref^X{[H+]_ref^Y[CH3COCH3]_ref^Z})
Since we are comparing trial B to itself, the ratio becomes:
1 = (k[I2]^X{[H+]^Y[CH3COCH3]^Z}) / (k[I2]_B^X{[H+]_B^Y[CH3COCH3]_B^Z})
Let's say the reference trial has values of [I2]_ref, [H+]_ref, and [CH3COCH3]_ref. We can use the information provided to find the values of X, Y, and Z.
Comparing the concentrations of trial B to the reference trial, we can calculate the exponents X, Y, and Z:
From Column 2: (10.0) / ([I2]_ref)^X = 1
From Column 3: (10.0) / ([H+]_ref)^Y = 1
From Column 4: (10.0) / ([CH3COCH3]_ref)^Z = 1
Since 10.0 divided by the concentration of each species gives us 1, we can conclude that X = Y = Z = 1.
Now, we can simplify the ratio equation to:
1 = k / (k[I2]_B{[H+]_B[CH3COCH3]_B})
Canceling the k terms:
1 = 1 / ([I2]_B{[H+]_B[CH3COCH3]_B})
Now, we can substitute the values from trial B into the equation:
1 = 1 / (10.0{10.0 * 10.0})
Simplifying the equation:
1 = 1 / (10.0 * 100.0)
1 = 1 / 1000
Therefore, the rate constant (k) at 25.0 degrees C, based on the data collected for trial B, is 0.001.
A 5.0 10.0 10.0 25.0 25.0 130
I copied the first row you had. I count 7 columns but you identify only 6.