Question

Reaction quotient: At a very high temperature, `K_c=65.0` for the following reversible reaction:
`2HI(g)hArr H_2(g)+I_2(g)`
The following concentrations were detected in a mixture.
`C_(HI)=0.50M`, `C_(H_2)=2.80M`, and `C_(I_2)=3.40M`
Is the system at equilibrium ? If not, in which direction must the reaction proceed for equilibrium to be established?

Answer 1

Strategy: First substitute these concentrations (which could be present if we started with a mixture of HI, `H_2`, and `I_2`) into the expression for the reaction quotient to calculate `Q_c`. Next we compare `Q_c` with the known value of `K_c` to see whether the system is at equilibrium and also to predict the direction of the reaction that leads to equilibrium.
Solution:
`Q_c=(C_(H_2)C_(I_2))/(C_(HI)^2)=((2.80)(3.40))/((0.50)^2)=38.1`
Since `Q_c` is smaller than `K_c(65.0)`, the system is not at equilibrium, i.e., `r_f!=r_b`. For equilibrium to be established, the value of `Q_c` must increase until it equals `K_c`. This can occur only if the numerator increases (i.e., an increase in the concentrations of `H_2` and `I_2`) and the denominator decreases (i.e., a decrease in the concentration of HI). Thus, the forward (left-to-right) reaction must occur to a greater extent than the reverse reaction `(r_f gt r_b)`, i.e., some HI must dissociate to form more `H_2` and `I_2` to reach equilibrium.