# Question 00c38

Jun 4, 2015

$p {H}_{2} O = 0.727 \text{Atm}$

$p {H}_{2} = 0.181 \text{Atm}$

$p {O}_{2} = 0.0909 \text{Atm}$

$K p = 0.00563 \text{Atm}$

$2 {H}_{2} {O}_{\left(g\right)} r i g h t \le f t h a r p \infty n s 2 {H}_{2 \left(g\right)} + {O}_{2 \left(g\right)}$

$\text{Initial moles}$:

$2. \ldots \ldots \ldots \ldots \ldots 0. \ldots \ldots \ldots \ldots \ldots . .0$

We are told that 20% of the steam dissociates. This means that 20% of 2 moles dissociate which = 2 x 0.2 = 0.4 moles.

This means that the number of moles of steam left @ equilibrium = (2 - 0.4) = 1.6 mol

From the equation we can see that if 0.4 mol steam dissociate then 0.4 mol of hydrogen must form as well as 0.2 mol oxygen.

So equilibrium moles $\Rightarrow$

$\left(2 - 0.4\right) \ldots \ldots \ldots .0 .4 \ldots \ldots \ldots . .0 .2$

Which is:

$1.6 \ldots \ldots \ldots \ldots \ldots 0.4 \ldots \ldots \ldots \ldots \ldots . .0 .2$

So the total moles @equilibrium $= 1.6 + 0.4 + 0.2 = 2.2$

We can now work out the mole fraction of each gas since

" mole fraction"=( "no. moles gas")/("total moles")

We can then find the partial pressure of each gas since for a gas $A$

pA="mol. fract A"xxp_("Total") $\Rightarrow$

$p {H}_{2} O = \frac{1.6}{2.2} \times 1 = 0.727 \text{Atm}$

$p {H}_{2} = \frac{0.4}{2.2} \times 1 = 0.181 \text{Atm}$

$p {O}_{2} = \frac{0.2}{2.2} \times 1 = 0.0909 \text{Atm}$

Lets do a check here and see if they add up to the total pressure of $1 \text{Atm}$:

$0.727 + 0.181 + 0.0909 = 0.999$ - so that looks ok.

Now for ${K}_{p}$:

${K}_{p} = \frac{{\left(p {H}_{2}\right)}^{2} \times p {O}_{2}}{{\left(p {H}_{2} O\right)}^{2}}$

$= \frac{{0.181}^{2} \times 0.0909}{{0.727}^{2}}$

${K}_{p} = 0.00563 \text{Atm}$

Jun 4, 2015

Here's how you should approach this problem.

You know that at a total pressure of 1 atm and at very high temperatures, water dissociates to form hydrogen gas, ${H}_{2}$, and oxygen gas, ${O}_{2}$.

The equilibrium reaction that describes that dissociation looks like this

$2 {H}_{2} {O}_{\left(g\right)} r i g h t \le f t h a r p \infty n s 2 {H}_{2 \left(g\right)} + {O}_{2 \left(g\right)}$

Notice that 2 moles of water produce 2 moles of hydrogen gas and 1 mole of oxygen.

You also know that only 20% of the molecules dissociate. If you have $x$ be the total number of molecules present, then you have

$\frac{20}{100} \cdot x \to$ the number of water molecules that dissociate;

$\frac{80}{100} \cdot x \to$ the number of water molecules that don't dissociate.

Out of the molecules that do dissociate, you know that each pair produces a pair of hydrogen molecules and 1 oxygen molecule. This means that you have

$\frac{20}{100} \cdot x \to$ the number of molecules of hydrogen produced;

$\frac{1}{2} \cdot \frac{20}{100} \cdot x \to$ the number of molecules of oxygen produced.

So, at equilibrium, your vessel will contain

$\frac{80}{100} \cdot x \to$ undissociated water molecules;

$\frac{20}{100} \cdot x \to$ hydrogen molecules;

$\frac{10}{100} \cdot x \to$ oxygen molecules.

The total number of molecules will be

${n}_{\text{total}} = \left(\frac{80}{100} + \frac{20}{100} + \frac{10}{100}\right) \cdot x = \frac{110}{100} \cdot x$

You can write the partial pressure of each gas by using the mole fraction they hold in the mixture and the total pressure of the mixture by using Dalton's Law of Partial Pressures.

${P}_{\text{component" = chi_"component" * P_"total}}$

The mole fraction is defined as the number of moles of a gas divided by the total number of moles of gas present in a mixture. In your case, you have

${\chi}_{\text{oxygen" = (10/100 * cancel(x))/(110/100 * cancel(x)) = 1/11}}$

${\chi}_{\text{hydrogen}} = \frac{\frac{20}{100} \cdot \cancel{x}}{\frac{110}{100} \cdot \cancel{x}} = \frac{2}{11}$

${\chi}_{\text{water}} = \frac{\frac{80}{100} \cdot \cancel{x}}{\frac{110}{100} \cdot \cancel{x}} = \frac{8}{11}$

The partial pressures of the three gases will be

P_"oxygen" = 1/11 * "1 atm" = color(green)(1/11 "atm")

P_"hydrogen" = 2/11 * "1 atm" = color(green)(2/11 "atm")

P_"water" = 8/11 * "1 atm" = color(green)(8/11 "atm")

The equilibrium constant, ${K}_{p}$, is defined as

${K}_{p} = \frac{{\left({H}_{2}\right)}^{2} \cdot \left({O}_{2}\right)}{{H}_{2} O} ^ 2$

K_p = ((2/11)^2 cancel("atm"^2) * 1/11 "atm")/((8/11)^2 cancel("atm"^2)) = 4/cancel(121) * 1/11 * cancel(121)/64 "atm"

K_p = 1/(11 * 16) "atm"= color(green)("0.00568 atm")#