# A mixture of .47 moles of H2 and 3.59 moles of HCl is heated to 2800'C. Calculate the equilibrium partial?

pressure of H2,Cl2,and HCl if the total pressure if 2.00 atm . For the reaction H2(g) + Cl2(g) -----> 2HCl(g)

Kp is 193 at 2800'C

If anyone please help me solve this and if you can please guide me stop by step, i'm really lost.

H2 + Cl2 <-----> 2 HCl

initial concentration

0.47. . . .3.59

at equilibrium

0.47-x . . 3.59-x . . . .2x

We have supposed V = 1.0 L

Since delta n = 0 ( 2 - 1 - 1 = 0 ) kp = kc

193 = (2x)^2 / (0.47 -x)(3.59-x)

Solve for x so you know moles HCl , H2 and Cl2 then the partial pressures

woah... that's confusing.

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Kp is 193 at 2800'C

If anyone please help me solve this and if you can please guide me stop by step, i'm really lost.

**Answer:**H2 + Cl2 <-----> 2 HCl

initial concentration

0.47. . . .3.59

at equilibrium

0.47-x . . 3.59-x . . . .2x

We have supposed V = 1.0 L

Since delta n = 0 ( 2 - 1 - 1 = 0 ) kp = kc

193 = (2x)^2 / (0.47 -x)(3.59-x)

Solve for x so you know moles HCl , H2 and Cl2 then the partial pressures

woah... that's confusing.

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