Question

Hello, I have tried to make a clear scan of my problems In a review I am nearly done with but I wanted to know if someone could walk me through these practice problem. I want to understand it better, so thank you for the help!

Use the following data at 1200 K to estimate a value of Kp for the reaction, 2H2(g) + O2(g) -2H20(g) C(graphite) + CO2(g) , 2C0(g) CO2(g)+H2(9)CO(g)+H20(g)Kc-1.4 C(graphite) + ½ O2(g)-CO(g) Kc=0.64 KC#1.0x108 At 25°C, the Kc for the following reaction is 8.96x10 PCls(g)PCl3(g)+Cl2(9) s a mixture of 6.70 mol of PCls, 0.308 mol of PCls, and 4.22 mol of Ch in a 1.50 L container at equilibrium? If not, which way will the reaction proceed?

Answer 1

Ans 1

Write the reverse of the first equation and multiply by 2

4CO = 2C + 2CO2

K1 = 1/0.64 = (1.5625)² = 2.4414

Write the second reaction and multiply by 2

2CO2 + 2H2 = 2 CO + 2H2O

K2 = 1.4² = 1.96

Write the third reaction and multiply by 2

2C + O2 = 2 CO

K3 = (1 x 10⁸)² = 10¹⁶

Now add all these three reactions

4CO + 2CO2 + 2H2 + 2C + O2 = 2C + 2CO2 + 2CO + 2H2O + 2CO

2H2 + O2 = 2 H2O

Overall Kc = K1 x K2 x K3

= 2.4414 x 1.96 x 10^16

= 4.79 x 10^16

Relation between Kp and Kc

Kp = Kc x (RT)^$\Delta$n

$\Delta$n = sum of stoichiometric coefficient of gaseous products - sum of stoichiometric coefficient of gaseous reactants

= 2 - (2 + 1) = - 1

Gas constant R = 0.0821 L-atm/mol-K

Temperature T = 1200 K

Kp = 4.79 x 10^16 x (0.0821 x 1200)^-1

= 4.86 x 10^14