The reaction CHCL3(g)+CL2(g)_CCL4(g)+HCL(g) has the following rate law:Rate =k [CHCL3][CL2].If the concentration of CHCL3 is increased by a factor of five while the concentration of CL2 Is kept the same use a mathematical equation to show the factor by which the rate will change.
Well, well, well, we have ourselves a little chemical equation, don't we? Let's see if I can clown around with some math to find the factor by which the rate will change.
We know that the rate law is given by Rate = k[CHCl3][Cl2]. And you're telling me that the concentration of CHCl3 is increased by a factor of five while the concentration of Cl2 remains the same. Now, let me put on my mathematical hat and help you out here.
If the concentration of CHCl3 is multiplied by a factor of five, we can say that [CHCl3]new = 5[CHCl3]old. Since the concentration of Cl2 remains the same, [Cl2]new = [Cl2]old.
Now we can substitute these values into the rate law equation to find the new rate:
Ratenew = k[CHCl3]new[Cl2]new
= k(5[CHCl3]old)([Cl2]old)
Now, we can simplify this expression:
Ratenew = 5k[CHCl3]old[Cl2]old
So, the new rate is equal to five times the old rate. Therefore, the factor by which the rate will change is 5.
Ta-da! Math and chemistry have come together to create a hilarious equation.
Based on the given rate law, Rate = k [CHCl3] [Cl2], if the concentration of CHCl3 is increased by a factor of five, the new concentration of CHCl3 will be 5[CHCl3].
To determine the factor by which the rate will change, we need to compare the rates before and after the concentration change.
Before concentration change: Rate1 = k [CHCl3] [Cl2]
After concentration change: Rate2 = k (5 [CHCl3]) [Cl2]
To find the factor by which the rate will change:
Rate2/Rate1 = (k (5 [CHCl3]) [Cl2]) / (k [CHCl3] [Cl2])
The k and [Cl2] terms will cancel out:
Rate2/Rate1 = (5 [CHCl3]) / [CHCl3]
Simplifying further:
Rate2/Rate1 = 5
This shows that the rate will increase by a factor of 5 when the concentration of CHCl3 is increased by a factor of 5 while the concentration of Cl2 is kept the same.
To determine the factor by which the rate will change when the concentration of CHCl3 is increased by a factor of five while the concentration of Cl2 is kept the same, we can use the rate law equation.
Given the rate law: Rate = k [CHCl3] [Cl2]
Let's consider the initial rate, which we'll call Rate1, when the concentrations of CHCl3 and Cl2 are at their original values, represented by [CHCl3]1 and [Cl2]1, respectively.
Rate1 = k [CHCl3]1 [Cl2]1
Now, let's consider the rate when the concentration of CHCl3 is increased by a factor of five. The concentration of CHCl3 after the increase will be represented by [CHCl3]2 = 5 * [CHCl3]1, while the concentration of Cl2 remains the same ([Cl2]2 = [Cl2]1).
Rate2 = k [CHCl3]2 [Cl2]2
Substituting the new concentrations into the rate law equation:
Rate2 = k (5 * [CHCl3]1) [Cl2]1
Rate2 = 5k [CHCl3]1 [Cl2]1
Comparing Rate2 and Rate1, we can determine the factor by which the rate changes:
Rate2 / Rate1 = (5k [CHCl3]1 [Cl2]1) / (k [CHCl3]1 [Cl2]1)
Rate2 / Rate1 = 5
Therefore, when the concentration of CHCl3 is increased by a factor of five while the concentration of Cl2 is kept the same, the rate will increase by a factor of five.
The rate will increase by a factor of 5. Since they want a math equation I would do this.
Make up a number for the rate for concns of 1 for each reactant and solve for k. That is
rate = k(CHCl3)(Cl2)
25 = k(1)(1)
k = 25/1 = 25
Then increase the CHCl3 by 5 but leave the Cl2 at 1 as per the problem.
rate = k(CHCl3)(Cl2)
rate = 25(5)(1) = 125
first rate = 25
second rate = 125
factor is 125/25 = 5