Consider the reaction:
2 O3( g)¡3 O2( g)
The rate law for this ...

Question

Consider the reaction: 
2 O3( g)¡3 O2( g) 
The rate law for this reaction is: 
Rate = k 
[O3]2 
[O2] 
Suppose that a 1.0-L reaction vessel initially contains 1.0 mol of 
O3 and 1.0 mol of O2. What fraction of the O3 will have reacted 
when the rate falls to one-half of its initial value?

Accepted Answer

Hi everyone here we have a question telling us that the rate law for the reaction A to B is ready equals K times concentration of a divided by concentration of B squared. An initial setup involves a 1.0 liter flask with a 0. mol of A and 0.5 mol of B determine the fraction of a that has reacted when the rate of the reaction is a third of the initial rate. So the concentration of A Equals 0.5 Molar. Because my polarity equals moles over leader and we have 0.5 moles per one liter. And the concentration of B Is also 0.5 malware. R right one equals k Times 0.5, Divided by 0.5 Squared Equals three. Rate too right to equals k Times 0. minus x divided by 0.5 Plus two x squared To get the ratio of right to and rate one. We are going to take rate too equals k Times 0.5 -X Divided by 0.5 Plus two x squared three rate too equals k Times 0. Divided by 0.5 squared one third Equals 0.5 More minus x Times 0. Squared Divided by 0. Plus two x squared 0.5 Plus two x Squared equals three times 0.5 - Times 0.5 Squared four X squared Plus 2.75 x -0.125 equals zero x equals zero point 0428. The fraction of a reacted Is 0.042, eight marbles, Divided by 0.5 moles Equals 0.086. And that is our final answer. Thank you for watching. Bye.

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