When 1.67 mol of NH 3 (g) was placed into a one Liter container at 350ºC, it was

Question

When 1.67 mol of NH3(g) was placed into a one Liter container at 350ºC, it was found to form 0.45 moles of N2(g) when the equilibrium shown below was reached:

                                                2 NH3(g) N2(g) + 3 H2(g)

                        a)         What are the H2(g) and NH3(g) concentrations at equilibrium?

[H2] = ________________                             [NH3] = _________________

                        b)         What is the value of the equilibrium constant, K, at 350ºC for this reaction?

Explanation / Answer

Number of mole = 1.67 mole
Volume = 1 L
So, concentration = 1.67 mol / 1 L = 1.67 M

2 NH3(g)       N2(g)    +    3 H2(g)
IC: 1.67   0 0
C: - 2x +x    +3x
EC: 1.67 - 2x    0.45 3x

So, x = 0.45

Hence, the equilibirum terms are

[N2(g)] = 0.45
[H2(g)] = 3x = 3 x 0.45 = 1.35
[NH3(g)] = 1.67 - 2x
      = 1.67 - 2(0.45)
      = 1.67 - 0.90
= 0.77

(b)

Equilibrium constant, K = { [N2(g)] x [H2(g)]3 } / [NH3(g)]2
   = { (0.45) x (1.35)3 } / (0.77)2
   = { (0.45) x (2.46) } / (0.59)
                             = 1.88

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.