Question

For the reaction N2O4(g) ⇋ 2NO2(g), Kc = 0.25 at 98°C. At a point during the reaction, the concentration of N2O4 = 0.50 M and the concentration of NO2 = 0.50 M. What is the value of Q? Enter a number to 2 decimal places. Is the reaction at equilibrium at that time? Enter either yes or no. In which direction is it progressing? Enter either right or left or at equilibrium.

Answer 1

Answer:

Q = 0.50

No

Left

Explanation:

At a generic reversible equation

aA + bB ⇄ cC + dD

The reaction coefficient (Q) is the ratio of the substances concentrations:

$Q = \frac{[C]^c*[D]^d}{[A]^a*[B]^b}$

Solids and liquid water are not considered in this calculus.

When the reaction achieves equilibrium (concentrations are constant), the Q value is named as Kc, which is the equilibrium constant of the reaction. If Q > Kc, it indicates that the concentration of the products is higher, so, the reaction must progress to the left and form more reactants; if Q < Kc, than the concentrations of the reactants, are higher, so, the reaction progress to the right.

In this case:

Q = $\frac{[NO_2]^2}{[N_2O_4]}$

$Q = \frac{0.50^2}{0.50}$

Q = 0.50

So, Q > Kc, the reaction is not at equilibrium and it progresses to the left.