Question 7bb00

Jul 2, 2014

At equilibrium, P_("CO"_2) = "305 Torr"

Explanation:

We start with the balanced chemical equation for the equilibrium.

${\text{CO" + "H"_2"O" ⇌ "CO"_2 +"H}}_{2}$

Then we write the ${K}_{P}$ expression.

${K}_{P} = \left({P}_{\text{CO"_2)P_("H"_2))/(P_("CO")P_("H"_2"O}}\right)$

Next, we set up an ICE table: Insert these values into the ${K}_{P}$ expression:

K_P = (P_("CO"_2)P_("H"_2))/(P_("CO")P_("H"_2"O")) = (x × x)/((1360-x)(1750-x)) = 0.0611

x^2/(2 380 000 – 3110x + x^2) = 0.0611

x^2 = 145 418 – 190.02x + 0.0611x^2

Putting all terms on the left, we get

${x}^{2} - 0.0611 {x}^{2} + 190.02 x - 145 418 = 0$

${x}^{2} \left(1 - 0.0611\right) + 190.02 x - 145 418 = 0$

$0.9389 {x}^{2} + 190.02 x - 145 418 = 0$

Use your calculator or an on-line quadratic equation solver such as the one at

$x = 305$

P_("CO") = (1360 – x)" Torr" = (1360 - 305)" Torr" = "1055 Torr"

P_("H"_2"O") = (1750 – x)" Torr" = (1750 – 305)" Torr" = "1445 Torr"

${P}_{C {O}_{2}} = {P}_{{H}_{2}} = x \text{ Torr" = "305 Torr}$

Check: (305 × 305)/(1055 × 1445) = 0.0610# (Close enough!)

The video below shows how to use an ICE table to solve a ${K}_{p}$ problem.