At a certain temperature, Kc = 0.914 for the reaction
NO2 (g) + NO (g) <----> N2O (g) + O2 (g)
Equal amounts of NO and NO2 are to be placed in a 5.00 L container until the N2O concentration at equilibrium is 0.050 M. How many moles of NO and NO2 must be placed in the container?
3 answers
the answer is 0.261 moles
..........NO2 + NO ==> N2O + O2
initial....x.....x.....0......0
equil................0.05....0.05
Kc = 0.914 = (N2O)(O2)/(NO2)(NO)
(N2O) = x
(NO) = x
(N2O) = 0.05
(O2) = 0.05
Solve for x which is molarity NO2 and NO.
Then moles = M x L to solve for moles.
initial....x.....x.....0......0
equil................0.05....0.05
Kc = 0.914 = (N2O)(O2)/(NO2)(NO)
(N2O) = x
(NO) = x
(N2O) = 0.05
(O2) = 0.05
Solve for x which is molarity NO2 and NO.
Then moles = M x L to solve for moles.
What is the value of equilibrium constant kc for NO2 + N2O =3NO
Given equilibrium concentrations are NO2=1.25,N2O=1.80 and NO=0.0015
Given equilibrium concentrations are NO2=1.25,N2O=1.80 and NO=0.0015
4 answers