Question

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, the rate will

Answer 1

We know that

v1 = k[CHCl3][Cl2]

If the concentration of CHCl3 is increased by a factor of five while the concentration of Cl2 is kept the same, then :

v2 = k[5CHCl3][Cl2]

v2 = 5(k[CHCl3][Cl2])

v2 = 5v1

The rate will increase five times