Question

at a given temperature, 0.300 mole NO, 0.200 mol Cl2 and 0.500mol ClNO were placed in a 25.0 liter container. the following equilibrium established:
2ClNO(g)<->2NO(g)+Cl2(g)
1.at equilibrium, 0.600 mol of ClNO was present. what is the number of moles of Cl2 present at equilibrium
2.The equilibrium constant

3. 2 mol of NO is placed in a 1 liter flask at 2273K. After equilibrium is attained, 0.0863 mol N2 and 0.0863 mol O2 are present. what is Kc for this reaction ?
2NO(g)<->N2(g)+O2(g)

Answer 1

K = NO^2 *Cl2 / (ClNO)^2

[NO] = 0.3/25 = 0.012

[Cl2] = 0.2/25 = 0.008

[ClNO] = 0.5/25= 0.02

in equilibrium

[NO] = 0.012 + 2x

[Cl2] = 0.008 + x

[ClNO] = 0.02 -2x

we know that

[ClNO] = 0.02 -2x = (0.6/25)

x = ((0.6/25)-0.02)/(-2) = -0.002

solv efor

[NO] = 0.012 + 2* -0.002 = 0.008

[Cl2] = 0.008 + -0.002 = 0.006

[ClNO] = 0.02 -2* -0.002 = 0.024

2

K = NO^2 *Cl2 / (ClNO)^2

K = (0.008^2)(0.006)/(0.024^2) = 0.0006666

3

when adding 2 mol of NO

[NO] = 2

[N2] = 0

[O2] = 0

in equilibrium

[NO] = 2 - 2x

[N2] = 0 +x = 0.0863

[O2] = 0+x = 0.0863

then

[NO] = 2 - 2*0.0863 = 1.8274

K = (N2)(O2) /(NO)^2 = (0.0863 )(0.0863 )/(1.8274^2)

K = 0.00223025411