# Consider the reaction Q+R→S+T and the rate law for the reaction rate:rate = k[Q]0[R]2.?

You run the reaction three times with [R] = 2.0M. For each run, you change the starting concentration of [Q] as follows: run 1, [Q] = 0.0M; run 2, [Q] = 1.0M; run 3, [Q] = 2.0M. Rank the rate of the three reactions using each of these concentrations.

Answer:
Since the rate law shows that the reaction is zero order for Q, the starting concentration of Q will not have an effect on the rate. The rates for all 3 runs will be the same, assuming you did not change the concentration of R.

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