Question #84ead
1 Answer
Whether the system is at equilibrium or not can be judged by the closeness of
They are both written almost the same way, but they use concentrations from different moments during the reaction (not at equilibrium or at equilibrium, respectively).
In this case, we are apparently not even close to equilibrium. It's way skewed towards the product.
(That's a good thing. Industrial synthesis processes try to push the reaction forward to minimize cost and maximize product yield!)
And the basic summary of this idea is:
- If
#Q < K# , the reaction shall shift to the right, and was previously skewed towards the reactants. - If
#Q > K# , the reaction shall shift to the left, and was previously skewed towards the products. - If
#Q = K# , we're good to go, the reaction is already at equilibrium...
You were given the reaction
#"CO"(g) + 2"H"_2(g) -> "CH"_3"OH"(g)#
And apparently, you were given the change in Gibbs' free energy at
#DeltaG_("700 K")^@ = "13.5 kJ/mol"#
This is not entirely clear from your question, so you should check your question later... If this is correct, however, recall the equation for relating
#color(green)(DeltaG = DeltaG^@ + RTlnQ)#
In this case, your reference temperature is
Either way,
#cancel(DeltaG)^(0) - RTlnK = DeltaG_"700 K"^@#
#=> DeltaG_"700 K"^@ = -RTlnK#
since
This gives you an equilibrium constant of:
#K_P = "exp"(-(DeltaG_"700 K"^@)/(RT))#
#= e^(-"13.5 kJ/mol"//"0.008314472 kJ/mol"cdot"K"//"700 K")#
#= ul(0.098)# (units omitted)
Now that you have an equilibrium constant to compare with, calculate the reaction quotient
#color(red)(Q_P) = (P_(CH_3OH))/(P_(CO)cdotP_(H_2)^2)#
#= (1 xx 10^(-6) "atm")/((2.0 xx 10^(-3) "atm")(1.0 xx 10^(-6) "atm")^2)#
#= 1/(2.0 xx 10^(-9) "atm"^2)#
#= ul(color(red)(5.0 xx 10^8))# (units omitted)
Here,