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.

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