K from [initial] and one [equilibrium]

Question

1)

A student ran the following reaction in the laboratory at 756 K:

N₂(g) + 3H₂(g) 2NH₃(g)

When she introduced 3.19×10^-2 moles of N₂(g) and 6.91×10^-2 moles of H₂(g) into a 1.00 liter container, she found the equilibrium concentration of H₂(g) to be 6.78×10^-2 M.

Calculate the equilibrium constant, K_c, she obtained for this reaction.

K_c =

2)

The equilibrium constant, K_c, for the following reaction is 1.24×10^-3 at 230 K.

2NOBr(g) 2NO(g) + Br₂(g)

When a sufficiently large sample of NOBr(g) is introduced into an evacuated vessel at 230 K, the equilibrium concentration of Br₂(g) is found to be 0.161 M.

Calculate the concentration of NOBr in the equilibrium mixture

Answer 1

N2(g) +  3H2(g) <===> 2NH3(g)

0.0319.....0.0691..............0.............Initial

-x............-3x...................+2x...........Change

0.0319-x...0.0691-3x........2x...........Equilibrium

0.0691-3x = 0.0678

x = 0.000433 moles

Final concentrations @ equilibrium:

[N2] = 0.0319 - 0.000433 = 0.0315 M

[H2] = 0.0678 M

[NH3] = 2 x 0.000433 = 0.000866 M

Kc = [NH3]² / [N2][H2]³

Kc = (0.000866)² / (0.0315)(0.0678)³

Kc = 0.0764

+++++++++++++++++++++++++++++++++++++++

2NOBr <====> 2NO + Br₂

Y........................0..........0...........Initial

-2x.....................+2x.......+x..........Change

Y-2x.................2x.........x...........Equilibrium

[Br₂] = x = 0.161 M

[NO] = 2x = 0.322 M

[NOBr] = Y - 0.322

Kc = [NO]²[Br₂] / [NOBr]²

1.24x10^-3 = (0.322)²(0.161) / (Y-0.322)²

1.24x10^-3 = 0.0167 / Y² - 0.644Y + 0.104

1.24x10^-3 Y² - 7.98x10^-4Y + 1.29x10^-4 = 0.0167

1.24x10^-3 Y² - 7.98x10^-4Y - 0.0166 = 0

y = 3.99

[NOBr]@ equilibrium = 3.99 - 0.322 = 3.67 M