Question #56b96

The reaction quotient will be equal to 0.718.

Here's how you can easily make a mistake when solving this problem. If you write the equilibrium like this

#C + H_2O rightleftharpoons CO + H_2#

you can be tempted to write the reaction quotient, #Q_c#, like this

#Q_c = ([CO] * [H_2])/([C] * [H_2O])#

You'd then go on to solve this equation and end up with #Q_c = 0.293#. You'd be wrong and your answer would be incorrect.

This is an excellent example of why you need to add the states of the species that take part in the reaction. Your equilibrium reaction actually looks like this

#C_((s)) + H_2O _((g)) rightleftharpoons CO_text((g]) + H_(2(g))#

Since carbon is in a solid state, its concentration is not a part of the expression for neither #Q_c#, nor #K_c#. This means that your reaction quotient would actually be

#Q_c = ([CO] * [H_2])/(cancel(color(red)([C])) * [H_2O]) = ([CO] * [H_2])/([H_2O])#

So, moving on with the explanation. The reaction quotient will tell you what the ratio between products and reactants is at a given moment during the reaction.

Think of #Q_c# as being a snapshot of how the amounts of reactants and products change as the reaction progresses towards equilibrium.

In your case, you know that at some point in the reaction, the reaction vessel contains that many moles of each species that takes part in the reaction.

Determine their concentration by using the volume of the reaction vessel

#C = n/V#

#[H_2O] = "12.9 moles"/"2.75 L" = "4.69 M"#

#[CO] = "3.00 moles"/"2.75 L" = "1.09 M"#

#[H_2] = "8.50 moles"/"2.75 L" = "3.09 M"#

Therefore, the reaction quotient will be

#Q_c = (1.09 * 3.09)/4.69 = color(green)(0.718)#

You'd then go on to compare this value with the value given to you for the equilibrium constant, #K_c#, and determine the direction in which the equilibrium is shifting.