Question

A) At what a certain temperature ,Kc equals 1.4x10^2 for the reaction:If a 3.00 L flask contains 0.400 mol CO2 and 0.100 mol of O2 at equilibrium, how many moles of CO are also present in thet certain flask?

B) A certain temperature the equilibrium constant, Kc equals 0.11 for the reaction: 2ICl(g)<>I2(g) +Cl2(g). What is the equilibrium concentration of ICl if 0.75 mol of I2 and 0.75 mol of Cl2 are initially mixed in a 2.0 L flask?

Answer 1

A) At what a certain temperature ,Kc equals 1.4x10^2 for the reaction:

If a 3.00 L flask contains 0.400 mol CO2 and 0.100 mol of O2 at equilibrium, how many moles of CO are also present in thet certain flask?

2 CO(g) + O2(g) ⇌ 2 CO2(g)

Kc = [CO2]^2/{[CO]^2x[O2]}

Now Kc = 1.4x10^2;

[CO2] = 0.400/3.00 = 0.133 M;

[O2] = 0.100/3.00 = 0.033 M

Substituting these values in the above equation, we get

1.4x10^2 = (0.133)^2/{[CO]^2x0.033}

[CO]^2 = 0.133^2/(0.033 x1.4x10^2) = 0.0038

[CO] = SQRT(0.0038) = 0.0616 M

Hence moles of CO = 0.0616 x 3.00 = 0.1848

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B) A certain temperature the equilibrium constant, Kc equals 0.11 for the reaction: 2ICl(g)<>I2(g) +Cl2(g). What is the equilibrium concentration of ICl if 0.75 mol of I2 and 0.75 mol of Cl2 are initially mixed in a 2.0 L flask?

2 ICl(g) ⇌ I2(g) + Cl2(g)
Kc = 0.11
Let at equilibrium concentration of ICl is X mol

OR

X/2.0M.

Then concentration of I2 and Cl2 = (0.75 - X) mol

Or (0.75 - X)/2.0 M

Hence

Kc = [I2][Cl2}/[ICl]^2

OR 0.11 = {(0.75 - X/2}x{(0.75 - X/2}/(X/2)^2
0.33 = (0.75 - X/2)/X/2
0.165X =0.75 - X/2

0.33X = 1.50 - X
1.33X = 1.50
X = 1.50/1.33 = 1.13 mol

So [ICl] = 1.13 mol/2.0L = 0.56M ------------answer