Question

Question #7bb00

Answer 1

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

Explanation:

We start with the balanced chemical equation for the equilibrium.

`"CO" + "H"_2"O" ⇌ "CO"_2 +"H"_2`

Then we write the `K_P` expression.

`K_P = (P_("CO"_2)P_("H"_2))/(P_("CO")P_("H"_2"O"))`

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 - 0.0611x^2 + 190.02x -145 418 = 0`

`x^2(1-0.0611) + 190.02x -145 418 = 0`

`0.9389x^2 + 190.02x -145 418 = 0`

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

http://www.math.com/students/calculators/source/quadratic.htm

`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_(CO_2) = P_(H_2) = x" 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.