Question

The reaction:

2 NO2(g) ↔ N2O4(g)

has an equilibrium constant, K_c, of 170 at 298K. Analysis of this system at 298K reveals that 4.57E-1 mol of NO₂ and 6.95E-2 mol of N₂O₄ are present in a 31.0-L flask. Determine the reaction quotient, Q, for this mixture. Enter your answer in scientific notation.

Which direction must the reaction in the previous question proceed to reach equilibrium. Note that you have only two tries for this question!

A. Right
B. Left

Answer 1

1)
[NO2] = mol of NO2 / volume in L
= 4.75*10^-1 mol / 31.0 L
= 1.53*10^-2 M

[N2O4] = mol of N2O4 / volume in L
= 6.95*10^-2 mol / 31.0 L
= 2.24*10^-3 M

Now use:
Qc = [N2O4]/[NO2]^2
= (2.24*10^-3) / (1.53*10^-2)^2
= 9.57

Answer: 9.57

2)
Qc = 9.57
Kc = 170
Qc is less than Kc.
So, reaction proceed in forward direction that is to the right
Answer: Right

Answer 2

Step 1: Calculate the Reaction Quotient (Q)

The reaction quotient $Q$ is calculated similarly to the equilibrium constant $K_c$, but uses the current concentrations instead of equilibrium concentrations.

Given Reaction:

$$2{\text{NO}}_2\text{(g)}\leftrightarrow {\text{N}}_2{\text{O}}_4\text{(g)}$$

Formula for Q:

$$Q=\frac{[{\text{N}}_2{\text{O}}_4]}{[{\text{NO}}_2{]}^2}$$

Given Data:

• Moles of ${\text{NO}}_2=4.57\times 10^{-1}\text{mol}$

• Moles of ${\text{N}}_2{\text{O}}_4=6.95\times 10^{-2}\text{mol}$

• Volume $V=31.0\text{L}$

Calculate Concentrations:

1. $[{\text{NO}}_2]=\frac{4.57\times 10^{-1}\text{mol}}{31.0\text{L}}=1.474\times 10^{-2}\text{M}$

2. $[{\text{N}}_2{\text{O}}_4]=\frac{6.95\times 10^{-2}\text{mol}}{31.0\text{L}}=2.242\times 10^{-3}\text{M}$

Compute Q:

$$Q=\frac{2.242\times 10^{-3}}{(1.474\times 10^{-2}{)}^2}=\frac{2.242\times 10^{-3}}{2.173\times 10^{-4}}=10.32$$

Scientific Notation:

$$Q=1.03\times 10^1$$

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Step 2: Compare Q to Kc to Determine Reaction Direction

• Given $K_c=170$ (equilibrium constant).

• Calculated $Q=10.32$.

Key Rule:

• If $Q<K_c$: Reaction proceeds right (toward products, ${\text{N}}_2{\text{O}}_4$).

• If $Q>K_c$: Reaction proceeds left (toward reactants, ${\text{NO}}_2$).

Here, $Q(10.32)<K_c(170)$, so the reaction shifts:
A. Right (to form more ${\text{N}}_2{\text{O}}_4$ until $Q=K_c$).

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Answers:

1. Reaction Quotient (Q): $1.03\times 10^1$

2. Direction to Reach Equilibrium: A. Right