Question

Question #44894

Answer 1

`K_p = 1.3 * 10^(-3)`

Explanation:

Start by taking a look at the given equilibrium reaction

`color(red)(2)"HI"_ ((g)) rightleftharpoons "H"_ (2(g)) + "I"_(2(g))`

Notice that the forward reaction reaction consists of `color(red)(2)` moles of hydrogen iodide, `"HI"`, reacting to form `1` mole of hydrogen gas, `"H"_2`, and `1` mole of iodine gas, `"I"_2`.

The position of this equilibrium will depend on the value of the equilibrium constant, `K_p`.

Now, you know that at equilibrium, the partial pressure of hydrogen iodide is `28` times that of hydrogen gas.

Since the stoichiometry of the reaction tells you that the reaction vessel must contain equal numbers of moles of hydrogen gas and iodine gas, it follows that the partial pressure of iodine gas must be equal to that of hydrogen gas.

This means that the partial pressure of hydrogen iodide is `28` times that of hydrogen gas and `28` times that of iodine gas. If you take `P_(HI)` to be the partial pressure of hydrogen iodide and `P` to be the partial pressure of hydrogen gas and of iodine gas, you can say that

`P_(HI) = 28 xx P`

By definition, the equilibrium constant for this gaseous equilibrium will look like this

`K_p = (P_(H_2) * P_(I_2))/P_(HI)^color(red)(2)`

This will be equivalent to

`K_p = (P * P)/(28P)^color(red)(2) = color(red)(cancel(color(black)(P^2)))/(28^2color(red)(cancel(color(black)(P^2)))) = 1/28^2 = color(green)(|bar(ul(color(white)(a/a)1.3 * 10^(-3)color(white)(a/a)|)))`

The result makes sense because the partial pressure of the reactant is significantly higher than that of the two products, which can only mean that he equilibrium lies further to the left.

In other words, at this temperature, the reaction vessel will contain more hydrogen iodide than hydrogen gas and iodine gas, which is why you have `K_p < 1`.