The equilibrium constant Kc for the reaction N2 (g) + 3H2 (g) -> 2NH3 (g)

at 450°C 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.

part 1: Enter the equilibrium concentration for N2.
part 2: Enter the equilibrium concentration for H2
part 3: Enter the equilibrium concentration for NH3.

[N2] = 0.3633M

[H2] = 1.090M

[NH3] = 0.2734M

Explanation:

Based on the reaction of the problem, Kc is defined as:

Kc = 0.159 = [NH3]² / [N2] [H2]³

Where [] are the equilibrium concentrations.

The initial concentrations of the reactants is:

N2 = 1.00mol / 2.00L = 0.500M

H2 = 3.00mol / 2.00L = 1.50M

When the equilibrium is reached, the concentrations are:

[N2] = 0.500M - X

[H2] = 1.50M - 3X

[NH3] = 2X

Where X is reaction quotient

Replacing in the Kc equation:

0.159 = [2X]² / [0.500 - X] [1.50 - 3X]³

0.159 = 4X² / 1.6875 - 13.5 X + 40.5 X² - 54 X³ + 27 X⁴

0.268313 - 2.1465 X + 6.4395 X² - 8.586 X³ + 4.293 X⁴ = 4X²

0.268313 - 2.1465 X + 2.4395 X² - 8.586 X³ + 4.293 X⁴ = 0

Solving for X:

X = 0.1367. Right solution.

X = 1.8286. False solution. Produce negative concentrations

Replacing:

[N2] = 0.500M - 0.1367M

[H2] = 1.50M - 3*0.1367M

[NH3] = 2*0.1367M

The equilibrium concentrations are:

43.452

step-by-step explanation:

284 times .153 = 43.452

the dissociation of nickel hydroxide is follows:

the concentration of hydroxide ion is

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

thus, concentration of ni is 0.23 * 10^-^4 mol /l