Question

CHCl3(g) + Cl2(g) → CCl4(g) + HCl(g) Rate = k[CHCl3]2[Cl2]3

If the concentration of CHCl3 is reduced to 1/3 of the initial amount and the concentration of Cl2 is tripled, what is the new rate going to be?

Answer 1

Let's start with some values. Say 3 M for CHCl3 and 3 M for Cl2 initially.
Then Rate = k[CHCl3]2[Cl2]3
rate = k*3^2*3^3 = 243 k.
Then 1/3 * 3 = 1 and 3^3 = 9
so new Rate = k[1^2 * 9^3] = 729 k. or without numbers we let x = (CHCl3) and y = (Cl2)
rate = k[x^2*y^3]. After changing these are
x/3 M and y^3 M
new rate = k(x/3)^2*(3y)^3 =k(x^2)/9 * (27y^3) = 3(x^2)(y^3)k.
Check all of that.