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

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 the rate of change of N2O5? Show your calculations.

Solutions

Expert Solution

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

queen_honey_blossom answered 1 year ago

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