For the equilibrium

$$\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)}$$

$$K_c = 245\ (\text{at}\ 1000\ \mathrm K)$$

The equilibrium concentrations are $[\ce{SO2}] = 0.102$, $[\ce{O2}] = 0.0132$, and $[\ce{SO3}] = 0.148$. Suppose that the concentration of $\ce{SO2}$ suddenly doubled. Calculate the $Q$ and use it to show that the forward reaction would take place to reach a new equilibrium.

I created an ice table so then I was able to find $x$:

$245=(0.148+2x)^2$ divided by $(0.0132-x)(0.204-2x)^2$


So the equilibrium concentrations are: $0.184$ for $\ce{SO2}$, $3.2\times10^{-3}$ for $\ce{O2}$ and $0.168$ for $\ce{SO3}$.
To check if the forward reaction is favored, $q<k$. The answer key states that $q=61.6$ but I don't know how they got it.


The reaction quotient, Q, is calculated the same way you would calculate the equilibrium constant, K. That is, Q is the product of the concentrations of the products of the reaction divided by the product of the concentration of the reactants,

$$ Q = \frac{[SO_3]^2}{[SO_2]^2 [O_2]} $$

Plugging the concentrations in $([SO_2] = 2 \times 0.102, [O_2] = 0.0132, [SO_3] = 0.148)$ should give you the answer for $Q$. I think you may have the concentrations written down incorrectly $([SO_3] = 0.184$ gives $Q = 61.6$, for instance$)$.

  • $\begingroup$ Why is the concentration of So2 multiplied by 2? $\endgroup$ – Shahad Jul 18 '16 at 19:08
  • $\begingroup$ Q=[SO3]^2 / [SO2]^2 [O2] 0.148^2 / 0.102^2 x 0.0132 = 8.134 not 61.6 $\endgroup$ – Shahad Jul 18 '16 at 19:09
  • $\begingroup$ Since Q is a measure of how far things are from equilibrium, you want to calculate Q based off of the non-equilibrium concentrations (i.e. immediately after the concentration of SO_2 is doubled.) $\endgroup$ – mmurbach Jul 18 '16 at 19:22
  • $\begingroup$ Oh ok but it still doesnt work. 0.2148^2 / 0.204^2 x 0.0132 = 39.87 not 61.6 $\endgroup$ – Shahad Jul 18 '16 at 19:26

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