In: 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 the rate of change of N2O5? Show your calculations.
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