# What is the rate law for this reaction?

My answer for this question is;

rate = k[A][C]^3

my calculations;

a = 2:2 = 1

b = 2:0 = 0

c = 3:9 = 3

• Welcome on Chemistry SE. You should provide your reasoning - why do you chose this answer. – Mithoron Oct 25 '15 at 16:35
• Check your calculation once again by putting the values. Hint: the rate quadruples when the concentration of reactant is doubled if $rate = k[reactant]^2$ – Chinmay Chandak Oct 25 '15 at 16:49
• Please type in your question when you get the time -- Images aren't searchable. – M.A.R. Oct 25 '15 at 17:10

$$\rm rate=k[A][C]^2$$
Taking 1 & 3 into account we find that on changing concentration of B while keeping A & C same, the rate remains same. So rate is independent of $\ce{[B]}$.
Then from 1 & 2 we find that rate is proportional to $\ce{[A]}$. Similarly, from 3 & 4 when concentration of C is made 3 times rate increases 9 times. So rate is proportional to $\ce{[C]^2}$.