Given the reaction...SO_2(g) + NO_2(g) rightleftharpoons SO_3(g) + NO(g)...for which K_"eq"=3.75, what will be the equilibrium concentrations of products and reactants?

Jan 22, 2017

$\left[S {O}_{3} \left(g\right)\right] = 0.680 + 0.216 = 0.896 \cdot m o l \cdot {L}^{-} 1$.

$\left[N O \left(g\right)\right] = 0.680 + 0.216 = 0.896 \cdot m o l \cdot {L}^{-} 1$.

$\left[S {O}_{2} \left(g\right)\right] = 0.680 - 0.216 = 0.464 \cdot m o l \cdot {L}^{-} 1$.

$\left[N {O}_{2} \left(g\right)\right] = 0.680 - 0.216 = 0.464 \cdot m o l \cdot {L}^{-} 1$.

Explanation:

The equilibrium follows the reaction:

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

And ${K}_{\text{eq}} = 3.75 = \frac{\left[S {O}_{3} \left(g\right)\right] \left[N O \left(g\right)\right]}{\left[S {O}_{2} \left(g\right)\right] \left[N {O}_{2} \left(g\right)\right]}$

And if $x \cdot m o l \cdot {L}^{-} 1$ $S {O}_{2}$ reacts.................

${K}_{\text{eq}} = 3.75 = \frac{\left(0.680 + x\right) \left(0.680 + x\right)}{\left(0.680 - x\right) \left(0.680 - x\right)}$

${K}_{\text{eq}} = 3.75 = \frac{{x}^{2} + 1.360 x + 0.462}{{x}^{2} - 1.360 x + 0.462}$

And so,

$3.75 {x}^{2} - 5.10 x + 1.73 = {x}^{2} + 1.360 x + 0.462$

And thus, $2.75 {x}^{2} - 6.46 x + 1.268 = 0$

This has roots at $x = 2.13 , \mathmr{and} x = 0.216$ (I used the quadratic equation for this solution!)

Clearly, the lower value is the only solution consistent with the starting conditions. And thus at equilibrium:

$\left[S {O}_{3} \left(g\right)\right] = 0.680 + 0.216 = 0.896 \cdot m o l \cdot {L}^{-} 1$.

$\left[N O \left(g\right)\right] = 0.680 + 0.216 = 0.896 \cdot m o l \cdot {L}^{-} 1$.

$\left[S {O}_{2} \left(g\right)\right] = 0.680 - 0.216 = 0.464 \cdot m o l \cdot {L}^{-} 1$.

$\left[N {O}_{2} \left(g\right)\right] = 0.680 - 0.216 = 0.464 \cdot m o l \cdot {L}^{-} 1$.

Just as a check, I reput these calculated values back into the equilibrium expression:

${\left(0.896\right)}^{2} / {\left(0.464\right)}^{2} \cong 3.75$ as required........

Good question; I am stealing it for my A2 class...........Gee, they will howl; especially as some of them will use $x = 2.13$.