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For the reaction 3O2(g) + 4NO(g)-->2N2O5(g) the rate of change in concentration of O2, delta[O2]/deltat = 0.057...

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Chemistry

For the reaction 3O2(g) + 4NO(g)-->2N2O5(g)

the rate of change in concentration of O2, delta[O2]/deltat =0.057 mol/L.sec. What is the rate of change of NO and what is therate of change of N2O5? Show your calculations.

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4.0 Ratings (802 Votes)

3O2(g) + 4NO(g) ----> 2N2O5(g)

Rate of the reaction = rate of disappearance of reactants = rate of appearance of products

   r = -(1/3)xd[O2]/dt = -(1/4)xd[NO]/dt = +(1/2)xd[N2O5]/dt        -----(1)

Given d[O2]/dt = 0.057 mol/L.sec

From Eqn(1) , -(1/3)xd[O2]/dt = -(1/4)xd[NO]/dt

                            d[NO]/dt = (4/3)xd[O2]/dt

                                          = (4/3) x 0.057

                                          = 0.076 mol/L.sec

-(1/3)xd[O2]/dt = +(1/2)xd[N2O5]/dt   

d[N2O5]/dt   = (2/3)xd[O2]/dt

                  = (2/3) x 0.057

                 = 0.038 mol/L.sec
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