At 603 K, Kc for the reaction NH4Cl <===> NH3 + HCl is 0.011. What is the equilibrium concentration of NH3 if the initial concentration of NH4Cl is 0.500 M. Please help me with this question because i got 2 answers so prove me.
Why not show your work for the two answers you have so we can find the error. That's a lot easier than us working the problem and letting you look for the error.
......NH4Cl(g) --> NH3(g) + HCl(g)
I......0.5..........0........0
C......-x...........x........x
E.....0.5-x.........x........x
Kc = 0.011 = (x)(x)/(0.50x)
Solve for x.
Please answer me with the working out...
please can you help me with the correct answer
To solve this question, we can use the equilibrium constant expression and an ICE table.
First, let's set up the equilibrium constant expression for the reaction NH4Cl <===> NH3 + HCl:
Kc = [NH3] * [HCl] / [NH4Cl]
Given that the value of Kc is 0.011, we can substitute this value in the equation:
0.011 = [NH3] * [HCl] / [NH4Cl] -------- Equation (1)
Now, let's set up an ICE table to track the changes in concentrations at equilibrium:
NH4Cl NH3 HCl
---------------------------------
Initial 0.500 M 0 M 0 M
Change -x x x
Equilibrium 0.500 - x x x
Now, substitute the values into the equilibrium constant expression (Equation 1) using the equilibrium concentrations:
0.011 = x * x / (0.500 - x)
Simplifying this equation, we get:
0.011 * (0.500 - x) = x^2
0.0055 - 0.011x = x^2
Rearranging the equation, we obtain:
x^2 + 0.011x - 0.0055 = 0
This is a quadratic equation in terms of x. We can solve it using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / 2a
In this case, a = 1, b = 0.011, and c = -0.0055. Substituting these values into the quadratic formula, we get:
x = (-0.011 ± √((0.011)^2 - 4(1)(-0.0055))) / (2(1))
Simplifying further, we have:
x = (-0.011 ± √(0.000121 + 0.022)) / 2
x = (-0.011 ± √0.022121) / 2
x = (-0.011 ± 0.1487) / 2
Simplifying the expression, we have two possible values for x:
x1 = (-0.011 + 0.1487) / 2
x1 = 0.06835
x2 = (-0.011 - 0.1487) / 2
x2 = -0.07935 (negative value, which is not valid since we are dealing with concentrations)
Therefore, the equilibrium concentration of NH3 is approximately 0.068 M.