Can the unit of rate of a reaction be s^-1?
I reacted different concentrations of glucose solutions with potassium permanganate and timed the reaction. Then I plotted time against concentration graph. Since I got a hyperbolic graph, I graphed 1/t against concentration to get a linear graph. Can I then say graph of rate against concentration? I'm a bit skeptical, because rate of reaction is change in concentration over time and in my case, I only have 1/time. So can I say rate, or do I leave it with 1/time?
Note: I know I could plot the rate of reaction (change in concentration over time) against concentration, but that's not what I'm supposed to do. It wouldn't help me with my further calculations.
Any help is appreciated.
2 Answers | Add Yours
This is an excellent question. You have plotted the glucose concentration versus the reciprocal of time (1/t) and gotten a straight line. You are right to be skeptical about calling 1/t the rate since a rate is change in concentration over time. We can think of the reciprocal of time as a sort of simplified stand in for the initial rate of the reaction since the initial rate is directly proportional to 1/t. So while you could call it the rate, it is probably best to keep it labelled as 1/t.
The importance of this graph is that is shows that the reaction is first order in respect to glucose. For a first order reaction, plotting rate (or in this case 1/t as a simplified stand in) versus concentration gives a straight line. Since the reaction is first order, you could also plot the natural logarithm (ln) of the concentration versus time to get a straight line. The slope of this line would be equal to -k (negative rate constant). Since the ln of a number is unitless, this means that the slope of the line would have the unit of 1/s, which is the correct unit for a first order rate constant.
Thank you for your answer. Though one thing still confuses me. If the graph 1/t against concentration is linear, shouldn't the reaction be second order and not first order?
We’ve answered 318,960 questions. We can answer yours, too.Ask a question