Asked by Sarah
2 NOCl(g) --> 2 NO(g) + Cl2(g)
Let’s assume that at a given temperature the equilibrium constant is 2.25: Also, the equilibrium concentration of the NOCl is 0.04M. Determine the concentration of the NO and the Cl2
CLUE: the concentration of both of the products must be equal if we started only from NOCl, since they are in a 1:1 ration. Since you don’t know the concentration of either of the products, use the variable ‘x’ to represent their concentration.
Let’s assume that at a given temperature the equilibrium constant is 2.25: Also, the equilibrium concentration of the NOCl is 0.04M. Determine the concentration of the NO and the Cl2
CLUE: the concentration of both of the products must be equal if we started only from NOCl, since they are in a 1:1 ration. Since you don’t know the concentration of either of the products, use the variable ‘x’ to represent their concentration.
Answers
Answered by
DrBob222
First, I don't buy that NO and Cl2 are equal and for exactly the reason the hint points out. The products are NOT a 1:1 ratio. (NO) must be twice the (Cl2).
.........2NOCl(g) --> 2NO(g) + Cl2(g)
equilib....0.04M.........2x.......x
K = 2.25 = (NO)^2(Cl2)/(NOCl)^2
2.25 = (2x)^2(x)/(0.04)^2
Solve for x = (Cl2), then 2x = (NO).
.........2NOCl(g) --> 2NO(g) + Cl2(g)
equilib....0.04M.........2x.......x
K = 2.25 = (NO)^2(Cl2)/(NOCl)^2
2.25 = (2x)^2(x)/(0.04)^2
Solve for x = (Cl2), then 2x = (NO).
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