Order of Reaction differs in equation and linearisation

I have given the following equation: $$A + 2B \rightarrow C$$

$$v = –k\cdot [A]\cdot[B]^2$$

gives me the orders 1 for A and 2 for B. That results in a total order of the reaction of 3. So far so good. But I have given the following plot for linearization:

But this linearization with $$\frac{1}{[A]}$$ defines a 2nd Order for A, not 1st.

The question is what does this mean for the given equation? And in which sequence could steps of the reaction occur?

• As this appears to be homework the rules here dictate that should add some thoughts about what is going on and what you have done to understand the data, otherwise the question may be closed without an answer. – porphyrin Feb 8 at 12:29
• This is an extremely interesting problem. Could you tell us more about this reaction ? What are the structure of A and B ? In which concentration ? How do you measure [A] ? Have you ever measured [B] ! – Maurice Feb 8 at 16:25
• @Maurice Unfortunately not, that's all the Information I have. But an idea: As the plot makes it clear, which order A must be, perhaps the reaction occurs as $2A + 4B \rightarrow 2C$. That could indicate that the given equation is not an elementary reaction and can be split up in (for example) following elementary reactions: $2B + A \rightarrow B_2A$ and then $2B_2A \rightarrow 2C$ . Then we would have one reaction with order 3 and one with order 1. stoichiometricly it would be the same as the given equation, but it couldn't happen that way. What do you think? – simande Feb 8 at 16:45