Consider the following reaction: (a) The rate law for this reaction is first ord
ID: 805854 • Letter: C
Question
Consider the following reaction:
(a) The rate law for this reaction is first order in S2O82-(aq) and first order in I-(aq). What is the rate law for this reaction?
Rate = k [S2O82-(aq)] [I-(aq)] Rate = k [S2O82-(aq)]2 [I-(aq)] Rate = k [S2O82-(aq)] [I-(aq)]2 Rate = k [S2O82-(aq)]2 [I-(aq)]2 Rate = k [S2O82-(aq)] [I-(aq)]3 Rate = k [S2O82-(aq)]4 [I-(aq)]
(b) If the rate constant for this reaction at a certain temperature is 0.00705, what is the reaction rate when [S2O82-(aq)] = 0.0332 M and [I-(aq)] = 0.0570 M?
Rate = M/s.
(c) What is the reaction rate when the concentration of S2O82-(aq) is doubled, to 0.0664 M while the concentration of I-(aq) is 0.0570 M?
Rate = M/s
Explanation / Answer
a)
Rate = k [S2O82-(aq)] [I-(aq)]
b)
Rate = k [S2O82-(aq)] [I-(aq)]
=0.00705 * 0.0332*0.0570
=0.00001334 M/s
c)
Rate = k [S2O82-(aq)] [I-(aq)]
=0.00705 * 0.0664*0.0570
=0.0000267 M/s