Question

for our equilibrium, Kc was calculated to be 4.6×10^-3 how did the current conditions. What does this imply for the concentrations of NO2 and N2O4? if you click on view image you can have a look at the cheat sheet containing the relevant equations.

Answer 1

The equilibrium constant (Kc) of 4.6×10^-3 suggests that under the current conditions there is a higher concentration of reactants (NO2) than products (N2O4) at equilibrium. However, exact concentrations cannot be calculated solely from the Kc value. Changes in temperature would also impact Kc and the equilibrium concentrations.

Step-by-step explanation:

The Kc value of 4.6×10^-3 that you calculated for your equilibrium relates directly to the concentrations of NO2 and N2O4 under the current conditions. Kc, or the equilibrium constant, describes the ratio of the concentrations of the products to reactants at equilibrium. In this case, considering the reaction represented by 2NO2 <--> N2O4, a low Kc implies there is a higher concentration of reactants (NO2) than products (N2O4) at equilibrium.

To understand this more precisely, consider the equation for Kc: Kc = [N2O4] / [NO2]^2. The low Kc suggests that the concentration of NO2 (in the denominator) is high and/or the concentration of N2O4 (in the numerator) is low. However, the specific concentrations of N2O4 and NO2 cannot be calculated only from the Kc value. This would require additional information relating initial and/or equilibrium concentrations or pressures.

Learn more about Chemical Equilibrium