The reaction: 2 NO2(g) ↔ N2O4(g) has an equilibrium constant, Kc, of 170 at 298K. Analysis of this system at 298K reveals that 1.02E-1 mol of NO2 and 8.00E-3 mol of N2O4 are present in a 27.0-L flask. Determine the reaction quotient, Q, for this mixture. Enter your answer in scientific notation.

The reaction quotient Q is 20.7

Explanation:

Step 1: Data given

Number of moles NO2 = 1.02 *10^-1 = 0.102 moles

Number of moles N2O4 = 8*10^-3 = 0.008 moles

Volume = 27.0 L

Step 2: The balanced equation

2 NO2(g) ↔ N2O4(g)

Step 3: Calculate concentrations

Concentration = moles / volume

[NO2] = moles NO2 / volume

[NO2] = 0.102 moles / 27.0 L

[NO2] = 0.00378 M

[N2O4] = moles N2O4 / volume

[N2O4] = 0.008 moles / 27.0L

[N2O4] = 2.96 *10^-4 M

Q = [N2O4] / [NO2]²

Q = (2.96 *10^-4) / (0.00378²)

Q = 20.7

The reaction quotient Q is 20.7

When Q<K, there are more reactants than products. As a result, some of the reactants will become products, causing the reaction to shift to the right.

a. energy is absorbed in the reaction.

if in a given reaction the energy of products is greater than the energy of reactants it means that reaction is endothermic and during the reaction, the energy is absorbed by reactants. in an endothermic reaction, the products have more stored chemical energy than the reactants. and in this reaction the reactants are more stable than products before the compounds with less energy are always more stable.