5.57

[Daniel Jeon Dis-2A]

I don't understand the logic for why the concentration of SO3 automatically becomes 0.245 - 0.240 = 0.005 mol. Can someone explain?

[Callie Phung 2J]

Hi Daniel,

For this question I set up an ICE table.

Since 0.245 moles of SO3 are already present in the vessel, 0.245 M is the initial concentration of SO3. We don't know how much NO is in the vessel yet, but we do know that there is no SO2 or NO2, so 0 is the initial concentration of SO2 and NO2.

Then, since everything is in a 1:1:1:1 stoichiometric ratio (the coefficients of the balanced reaction are all 1), the changes in all the concentrations will be plus or minus 1x. So SO3 will be 0.245 - x, NO will be n - x, and SO2 and NO2 will be + x.

But because we know the equilibrium concentration of SO2 will be 0.240M, then we can assume that x = 0.240 because we know that there was no SO2 to begin with in the initial concentration.

Therefore, in order to find how many moles of SO3 are reacting at equilibrium, we use its equilibrium concentration 0.245-x, but now substituting 0.240 for x: 0.245 - 0.240 = 0.005 mol SO3. Then you can use the K constant and equation, plugging in 0.005 as the SO3 equilibrium concentration and n - 0.240 as the NO equilibrium concentration, and then solve for n as your initial NO concentration.

Hope that helps.