Question

Equilibrium pressures question?

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 ?

Answer 1

`P_(H_2) = "0.0402 atm"`
`P_(I_2,eq) = "0.0442 atm"`
`P_(HI,eq) = "0.4216 atm"`

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Well, from your notes, you seem to have `K_p = 100` (in implied units of `"atm"`). So, let's set up the 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`,

`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; `K_p` is not small.

`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 `x = 0.0392` seems to work.

`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")`

Answer 2

use an ICE table, then find pressures with `K_p=100`?

Explanation:

My notes

Classmate's notes