At 603K,Kc for the reaction NH4Cl(g)<==>NH3(g) + HCl(g) is 0.011.What is the equilibrium concentration of NH3(g) if the initial concentration of NH4Cl(g) is 0.500M?
.................NH4Cl ==> NH3 + HCl
I................0.5 M............0..........0
C...............-x..................x...........x
E...............0.5-x.............x............x
Kc = 0.011 = (NH3)(HCl)/NH4Cl
Plug the E line into the Kc expression and solve for x = (NH3).
Post your work if you get stuck.
To find the equilibrium concentration of NH3(g), we can use the equation for the equilibrium constant (Kc) and the information given.
The given equilibrium equation is: NH4Cl(g) ⇌ NH3(g) + HCl(g)
The equilibrium constant expression for this reaction is:
Kc = [NH3(g)][HCl(g)] / [NH4Cl(g)]
Given that the equilibrium constant (Kc) is 0.011, and the initial concentration of NH4Cl(g) is 0.500 M, we need to find the concentration of NH3(g) at equilibrium.
Let's assume that the equilibrium concentration of NH3(g) is x M. Then, the equilibrium concentration of HCl(g) will also be x M since they have a 1:1 stoichiometric ratio.
Substituting the given values into the equilibrium constant expression:
0.011 = (x)(x) / (0.500)
Solving for x:
0.011 * 0.500 = x^2
0.0055 = x^2
Taking the square root of both sides:
x ≈ √0.0055
x ≈ 0.074 M
Therefore, the equilibrium concentration of NH3(g) is approximately 0.074 M.
To find the equilibrium concentration of NH3(g), we can use the equation:
Kc = [NH3(g)]/[NH4Cl(g)] * [HCl(g)]
In this case, the initial concentration of NH4Cl(g) is given as 0.500M.
Let's assume the equilibrium concentration of NH3(g) is x, which means the equilibrium concentration of NH4Cl(g) and HCl(g) is also x.
Now, we can substitute these values into the equation:
0.011 = x / 0.500 * x
Simplifying the equation:
0.011 = (x^2) / 0.500
Now, cross multiply:
0.011 * 0.500 = x^2
0.0055 = x^2
Taking the square root of both sides:
x = √0.0055
x ≈ 0.074 (to three decimal places)
Therefore, the equilibrium concentration of NH3(g) is approximately 0.074 M.