Asked by j
What is the initial rate of *appearance* of SO3(g)?
a reaction and table are given:
reaction: 2S02(g)+O2 -------> 2 SO3(g)
Table note: i= initial, IRD= initial rate of disappearance; concentrations given in molarity (M) IRD given in M/s.
Experiment [SO2]i [O2]i IRD O2
1 0.05 0.05 7.3*[10^(-2)]
2 0.10 0.05 2.9*[10^(-2)]
3 0.10 0.15 2.6*[10^(-1)]
4 0.12 0.12 ?????????????
so overall oreder is 4 or specifically:
rate=k([SO2]^2)([O2]^2) with k= 1168 so
rate=1168([SO2]^2)([O2]^2)
not sure what to do here I believe that rate relationship is:
-(delta [O2)]/delta time)=
1/2(delta [SO3]/delta time)
and I think I need to use trials 2&3 because that is where [SO2] is held constant.
Answer is given as 4.84*10^(-1) trying to figure out how to get there.
Thank you
a reaction and table are given:
reaction: 2S02(g)+O2 -------> 2 SO3(g)
Table note: i= initial, IRD= initial rate of disappearance; concentrations given in molarity (M) IRD given in M/s.
Experiment [SO2]i [O2]i IRD O2
1 0.05 0.05 7.3*[10^(-2)]
2 0.10 0.05 2.9*[10^(-2)]
3 0.10 0.15 2.6*[10^(-1)]
4 0.12 0.12 ?????????????
so overall oreder is 4 or specifically:
rate=k([SO2]^2)([O2]^2) with k= 1168 so
rate=1168([SO2]^2)([O2]^2)
not sure what to do here I believe that rate relationship is:
-(delta [O2)]/delta time)=
1/2(delta [SO3]/delta time)
and I think I need to use trials 2&3 because that is where [SO2] is held constant.
Answer is given as 4.84*10^(-1) trying to figure out how to get there.
Thank you
Answers
Answered by
DrBob222
Are you sure of these data? I can confirm that the reaction is second order with respect to O2 but I can't confirm it is second order with respect to SO2
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