# Equilibrium pressures question?

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If #5.00*10^-1# atm of #HI# gas, #0.00100# atm of #H_2# gas, and #0.00500# atm of #I_2# gas are present in a #5.00# L flask, calculate the equilibrium pressures of all 3 gases once equilibrium is established ?

If

##### 2 Answers

#P_(H_2) = "0.0402 atm"#

#P_(I_2,eq) = "0.0442 atm"#

#P_(HI,eq) = "0.4216 atm"#

Well, from your notes, you seem to have **reaction** and **ICE table** based solely on the numbers you have given here:

#"H"_2(g) " "" "+" "" " "I"_2(g)" " rightleftharpoons " " 2"HI"(g)#

#"I"" "0.00100" "" "" "0.00500" "" "" "0.500#

#"C"" "+x" "" "" "" "+x" "" "" "" "-2x#

#"E"" "0.00100+x" "0.00500+x" "0.500-2x#

Here, we have written that the reaction goes in reverse. Now that you have

#K_p = 100 = P_(HI)^2/(P_(H_2)P_(I_2))#

#= (0.500 - 2x)^2/((0.00100 + x)(0.00500 + x))#

Not much we can do here other than solve it in full;

#100 = (0.500 - 2x)^2/(5 xx 10^(-6) + 0.00600x + x^2)#

Multiply through by the denominator.

#5 xx 10^(-4) + 0.600x + 100x^2 = (0.500 - 2x)^2#

Expand the right-hand side.

#5 xx 10^(-4) + 0.600x + 100x^2 = 0.250 - 2 cdot 0.500 cdot 2x + 4x^2#

Get this into standard quadratic form:

#-0.2495 + 0.600x + 100x^2 = -2x + 4x^2#

#96x^2 +2.600x - 0.2495 = 0#

Now,

#a = 96#

#b= 2.600#

#c = -0.2495#

so that

#x = (-b pm sqrt(b^2 - 4ac))/(2a)#

#= (-(2.600) pm sqrt(2.600^2 - 4(96)(-0.2495)))/(2(96))#

This, as-given, has:

#x = 0.0392, -0.0662#

When we plug these in, however, only

#color(blue)(P_(H_2) = 0.00100 + (0.0392) = "0.0402 atm")#

#color(blue)(P_(I_2,eq) = 0.00500 + (0.0392) = "0.0442 atm")#

#color(blue)(P_(HI,eq) = 0.500 - 2(0.0392) = "0.4216 atm")#

use an ICE table, then find pressures with

#### Explanation:

**My notes**

**Classmate's notes**