Someone would be kind enough to help me with this exercise: 2"SO"_3(g) -> 2"SO"_2(g) + "O"_2(g) ?

Someone would be kind enough to help me with this exercise: At 1500 k they are introduced at a constant volume of 300 torr and 150 torr of SO3, SO2. It establishes the following equilibrium: $2 S {O}_{3} \to 2 S {O}_{2} + {O}_{2}$ to the equilibrium pressure that is 550 torr. Calculate kp and kc. Answers: Kp: 1.61 atm Kc: 1.31 x 10-2 pls help me fast . please show the working. how these answers came i dont understand there can be mistakes in the question . i think it is incomplete . what is incomplete

Jan 27, 2017

The gaseous reversible reaction under consideration at 1500K is:

$2 S {O}_{3} \left(g\right) r i g h t \le f t h a r p \infty n s 2 S {O}_{2} \left(g\right) + {O}_{2} \left(g\right)$

Here it is also given that $S {O}_{3} \left(g\right) \mathmr{and} S {O}_{2} \left(g\right)$ are introduced at a constant volume of 300 torr and 150 torr respectively. Since the pressure of a gas is proportional to the number of moles when their volume and temperature are constant. So we can say that the ratio of number of moles of $S {O}_{3} \left(g\right) \mathmr{and} S {O}_{2} \left(g\right)$ introduced is $300 : 150 = 2 : 1$. Let these are $2 x$ mol and $x$ mol

Now write the ICE table

$\textcolor{b l u e}{2 S {O}_{3} \left(g\right) \text{ "" "rightleftharpoons" "2SO_2(g)" "+" } {O}_{2} \left(g\right)}$

$\textcolor{red}{I} \text{ "2x" "mol" "" "" "" "" "x" mol"" "" "" "" "0" mol}$

$\textcolor{red}{C} - 2 \alpha x \text{ "mol" "" "+2alphax" mol"" "" "" "alphax" mol}$

$\textcolor{red}{E} \text{ "(1-alpha)2x" "mol" "(1+2alpha)x" mol"" "" "" "alphax" mol}$

where $\alpha$ represents the degree of dissociation at 1500K

So at equilibrium total number of moles of component gases in the reaction mixture is $\left(2 - 2 \alpha + 1 + 2 \alpha + \alpha\right) x = \left(3 + \alpha\right) x$
It is also given that at equilibrium the pressure of the reacton mixture is $550 \text{torr}$.

Now ratio of the total pressure with the initial pressure of $S {O}_{2} \left(g\right)$ should be equal to the ratio of their respective number of moles.

So $\left(550 \text{tor")/(150"tor}\right) = \frac{\left(3 + \alpha\right) x}{x}$

$\implies \alpha + 3 = \frac{11}{3}$
$\implies \alpha = \frac{11}{3} - 3 = \frac{2}{3}$

Now calculating mole fraction of the component gases at equilibrium

${\chi}_{S {O}_{3} \left(g\right)} = \frac{\left(1 - \alpha\right) 2 x}{\left(3 + \alpha\right) x} = \frac{\left(1 - \frac{2}{3}\right) 2}{\left(3 + \frac{2}{3}\right)} = \frac{2}{11}$

${\chi}_{S {O}_{2} \left(g\right)} = \frac{\left(1 + 2 \alpha\right) x}{\left(3 + \alpha\right) x} = \frac{1 + \frac{4}{3}}{\left(3 + \frac{2}{3}\right)} = \frac{7}{11}$

${\chi}_{{O}_{2} \left(g\right)} = \frac{\alpha x}{\left(3 + \alpha\right) x} = \frac{\frac{2}{3}}{\left(3 + \frac{2}{3}\right)} = \frac{2}{11}$

If P be the total pressure of the reaction mixture at equilibrium the the partial pressures of component gases will be

${p}_{S {O}_{3} \left(g\right)} = {\chi}_{S {O}_{3} \left(g\right)} \times P = \frac{2 P}{11}$

${p}_{S {O}_{2} \left(g\right)} = {\chi}_{S {O}_{2} \left(g\right)} \times P = \frac{7 P}{11}$

${p}_{{O}_{2} \left(g\right)} = {\chi}_{{O}_{2} \left(g\right)} \times P = \frac{2 P}{11}$

Now calculation of $\textcolor{red}{{K}_{p}}$

${K}_{p} = \frac{{p}_{S {O}_{2} \left(g\right)}^{2} \times {p}_{{O}_{2} \left(g\right)}}{{p}_{S {O}_{3} \left(g\right)}^{2}} = \frac{{\left(\frac{7 P}{11}\right)}^{2} \times \frac{2 P}{11}}{\frac{2 P}{11}} ^ 2$

$\implies {K}_{p} = \frac{49 P}{22}$

But given value of $P = 550 \text{ torr} = \frac{550}{760} a t m = \frac{55}{76} a t m$

So $\implies {K}_{p} = \frac{49 \times 55}{22 \times 76} \approx 1.61 a t m$

Now calculation of $\textcolor{b l u e}{{K}_{c}}$

We know the relation

$\textcolor{g r e e n}{{K}_{p} = {K}_{c} {\left(R T\right)}^{\Delta n}}$

where $\Delta n = \text{total number of moles of product gases"-"total number of moles of reactant gases}$

$\implies \Delta n = \left(2 + 1\right) - 2 = 1$

So ${K}_{c} = {K}_{p} / \left(R T\right)$

Here $R = 0.082 L a t m {K}^{-} 1 m o {l}^{-} 1$

And $T = 1500 K$

Inserting these values we get

$\textcolor{b l u e}{{K}_{c}} = \frac{1.61}{0.082 \times 1500} = 1.31 \times {10}^{-} 2$

Jan 27, 2017

Here's another way to do it. Your reaction was:

$2 {\text{SO"_3(g) rightleftharpoons 2"SO"_2(g) + "O}}_{2} \left(g\right)$

Since you have a constant volume, and since the temperature is assumed constant as well (as you are not given two temperatures), you can expect that the change in mols of gas relates mainly with the change in pressure, meaning that

$P = {P}_{1} + {P}_{2} + . . .$, Dalton's Law of Partial Pressures,

applies, and the given equilibrium pressure is the total pressure of all gases in the mixture.

Filling out an ICE Table gives:

${\text{ "" "" "2"SO"_3(g) rightleftharpoons 2"SO"_2(g)" " + " ""O}}_{2} \left(g\right)$

$\text{I"" "" ""300 torr"" "" ""150 torr"" "" "" "" ""0 torr}$
$\text{C"" "" ""-2x torr"" "" ""+2x torr"" "" "" ""+x torr}$
$\text{E"" "" ""300-2x torr"" ""150+2x torr"" "" ""x torr}$

Remember that the change in pressure will include the stoichiometric coefficients in front of the molecule in the balanced reaction.

But since you know that the equilibrium pressure was $\text{550 torr}$, you can use Dalton's law of partial pressures:

$P = \left(300 - 2 x\right) + \left(150 + 2 x\right) + x = 550$

$P = 450 + x = 550$

$\textcolor{g r e e n}{x = \text{100 torr}}$

That gives you each equilibrium partial pressure as:

${P}_{S {O}_{3}} = 300 - 2 \left(100\right) = \text{100 torr}$
${P}_{S {O}_{2}} = 150 + 2 \left(100\right) = \text{350 torr}$
${P}_{{O}_{2}} = \text{100 torr}$

Note that if you get a negative pressure, it means you mixed up the partial pressures of ${\text{SO}}_{2}$ and ${\text{SO}}_{3}$. if you don't get the right ${K}_{P}$, it may also be because your stoichiometric coefficients weren't incorporated into the ${K}_{P}$ expression.

The $\textcolor{b l u e}{{K}_{P}}$ is then:

${K}_{P} = \frac{{P}_{S {O}_{2}}^{2} {P}_{{O}_{2}}}{{P}_{S {O}_{3}}^{2}}$

$= \left({\left(\text{350 torr")^2("100 torr"))/(("100 torr}\right)}^{2}\right)$

$=$ $\text{1225 torr}$

Convert to $\text{atm}$ by dividing by $\text{760 torr/atm}$ to get $\textcolor{b l u e}{\text{1.6118 atm}}$.

Recall that ${K}_{P} = {K}_{C} \cdot {\left(R T\right)}^{\Delta {n}_{\text{gas}}}$. Since the mols of gas changed from 2 to 2+1 = 3, we say that $\Delta {n}_{\text{gas}} = 1$. Therefore:

color(blue)(K_C) = ("1.61 atm")/(("0.082057 L"cdot"atm/mol"cdot"K")("1500 K"))

$= 0.013095 = \textcolor{b l u e}{1.31 \times {10}^{- 2} \text{mol/L}}$

though it tends to be reported without units. Hope that helps!