Questions: Question 8 A reaction vessel contained 0.100 M N2 and O2 initially, determine the equilibrium concentration for nitrogen monoxide in the equilibrium below. N2 + O2 ↔ 2 NO Kp=0.100 0.100 0.0272 none of the above 0.0136 0.0316

Question 8
A reaction vessel contained 0.100 M N2 and O2 initially, determine the equilibrium concentration for nitrogen monoxide in the equilibrium below.
N2 + O2 ↔ 2 NO Kp=0.100
0.100
0.0272
none of the above
0.0136
0.0316

Transcript text: Question 8 A reaction vessel contained $0.100 \mathrm{M} \mathrm{N}_{2}$ and $\mathrm{O}_{2}$ initially, determine the equilibrium concentration for nitrogen monoxide in the equilibrium below. \[ \mathrm{N}_{2}+\mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} K_{p}=0.100 \] 0.100 0.0272 none of the above 0.0136 0.0316

Solution

Solution Steps

Step 1: Write the Equilibrium Expression

For the given reaction: \[ \mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2 \mathrm{NO} \] The equilibrium constant expression in terms of partial pressures is: \[ K_p = \frac{P_{\mathrm{NO}}^2}{P_{\mathrm{N}_2} P_{\mathrm{O}_2}} \] Given $ K_p = 0.100 $.

Step 2: Relate Concentrations to Partial Pressures

Assuming ideal gas behavior, the partial pressures can be related to molar concentrations using the ideal gas law. However, since we are given initial concentrations and not partial pressures, we will use the concentrations directly in the equilibrium expression.

Step 3: Set Up the ICE Table

Let $ x $ be the change in concentration of $\mathrm{NO}$ at equilibrium. The initial concentrations are: \[ [\mathrm{N}_2] = 0.100 \, \mathrm{M}, \quad [\mathrm{O}_2] = 0.100 \, \mathrm{M}, \quad [\mathrm{NO}] = 0 \, \mathrm{M} \] At equilibrium, the concentrations will be: \[ [\mathrm{N}_2] = 0.100 - x, \quad [\mathrm{O}_2] = 0.100 - x, \quad [\mathrm{NO}] = 2x \]

Step 4: Substitute into the Equilibrium Expression

Substitute the equilibrium concentrations into the equilibrium expression: \[ K_p = \frac{(2x)^2}{(0.100 - x)(0.100 - x)} = 0.100 \] Simplify the equation: \[ 0.100 = \frac{4x^2}{(0.100 - x)^2} \]

Step 5: Solve for $ x $

Take the square root of both sides: \[ \sqrt{0.100} = \frac{2x}{0.100 - x} \] \[ 0.3162 = \frac{2x}{0.100 - x} \] Rearrange to solve for $ x $: \[ 0.3162 (0.100 - x) = 2x \] \[ 0.03162 - 0.3162x = 2x \] \[ 0.03162 = 2.3162x \] \[ x = \frac{0.03162}{2.3162} \approx 0.0136 \]

Step 6: Calculate the Equilibrium Concentration of $\mathrm{NO}$

The equilibrium concentration of $\mathrm{NO}$ is: \[ [\mathrm{NO}] = 2x = 2 \times 0.0136 = 0.0272 \, \mathrm{M} \]