Asked by Anonymous
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?
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?
Answers
Answered by
Anonymous
the answer is 0.261 moles
Answered by
DrBob222
..........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.
Answered by
FREDLY
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
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