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.

Answer 1

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