The equilibrium constant kc for the reaction
N2(g)+ 3H(g)---2NH3(g)
at 450C is 0.159. Calculate the equilibrium composition when 1.00 mol N2 is mixed with 3.00 mol H2 in a 2.00-L vessel.
need help setting up the problem
Correct the equation to
N2 + 3H2 ==> 2NH3
Set up an ICE chart.
initial:
N2 = 1 mole/2 L = 0.5 M
H2 = 3 moles/2 L = 1.5 M
NH3 = 0
change:
NH3 = +2x
N2 = -x
H2 = -3x
equilibrium:
N2 = 0.5-x
H2 = 1.5-3x
NH3 = 2x
Substitute into Kc expression and solve. Post your work if you get stuck.
To calculate the equilibrium composition of the reaction, we need to use the given equilibrium constant (Kc) and the initial amounts of reactants.
First, let's determine the initial number of moles for each species in the reaction:
Initial number of moles of N2 = 1.00 mol
Initial number of moles of H2 = 3.00 mol
Initial number of moles of NH3 = 0 mol
Next, we need to set up an expression for the change in the number of moles for each species. In this case, since the stoichiometry of the reaction is given, we can determine the changes as follows:
Change in moles of N2 = -x (since 1 mole of N2 reacts to form 2 moles of NH3)
Change in moles of H2 = -3x (since 3 moles of H2 react to form 2 moles of NH3)
Change in moles of NH3 = +2x (since 2 moles of NH3 are produced)
Finally, we can use the equilibrium constant equation to set up an expression for Kc:
Kc = [NH3]^2 / ([N2] * [H2]^3)
Since the initial concentration of NH3 is 0, we can rewrite the expression as:
Kc = (2x)^2 / ([1.00 - x] * [3.00 - 3x]^3)
Now, we can substitute the given value for Kc (0.159) and solve for x. Once we find the value of x, we can use it to calculate the equilibrium composition by substituting it back into the expressions for the changes in moles.