At 27 C, Kp=1.5e18 for the rxn: 3 NO (g) <--> N2O (g) + NO2 (g) If 0.03 mol of N
ID: 927010 • Letter: A
Question
At 27 C, Kp=1.5e18 for the rxn:3 NO (g) <--> N2O (g) + NO2 (g)
If 0.03 mol of NO were placed in a 1 liter vessel and equilibrium were established, what would be the equilibrium concentrations of the products and reactants? At 27 C, Kp=1.5e18 for the rxn:
3 NO (g) <--> N2O (g) + NO2 (g)
If 0.03 mol of NO were placed in a 1 liter vessel and equilibrium were established, what would be the equilibrium concentrations of the products and reactants?
3 NO (g) <--> N2O (g) + NO2 (g)
If 0.03 mol of NO were placed in a 1 liter vessel and equilibrium were established, what would be the equilibrium concentrations of the products and reactants?
Explanation / Answer
Given equilibrium constant , Kp = 1.5x1018
We know that Kp = Kc x (RT)n
Where
n = change in number of moles
= total number of gaseous products - total number of gaseous reactants
= (1+1) - 3
= -1
R = gas constant = 0.0821 Latm . (mol-K)
T = temperature = 27 oC = 27+273 = 300 K
Plug the values we get Kc = Kpx (RT)-n
= (1.5x1018) x (0.0821x300)1
= 6.1 x1016
Initial concentration of NO is = number of moles / volume in L
= 0.03 mol / 1 L
= 0.03 M
3 NO (g) <--> N2O (g) + NO2 (g)
initial conc 0.03 0 0
change -3a +a +a
Equb conc 0.03-3a a a
Equilibrium constant , Kc = ([NO2][N2O]) / [NO]2
6.1 x1016 = ( a x a ) / (0.03-3a)2
On solving we get a = 0.01M
Therefore the equilibrium concentration of N2O = a = 0.01 M
equilibrium concentration of NO2 = a = 0.01 M
equilibrium concentration of NO = 0.03-3a = 0.03-(3x0.01) = 0.0 M
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