Multiple Choice
If Kc for the following reaction is 5.4 × 1013,
2 NO (g) + O2 (g) ⇆ 2 NO2 (g)
what is Kc for the reaction shown below?
4 NO2 (g) ⇆ 4 NO (g) + 2 O2 (g)
16. Chemical Equilibrium
ICE Charts
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At a given temperature the gas phase reaction: N2 (g) + O2 (g) ⇄ 2 NO (g) has an equilibrium constant of 4.18 x 10-7. What will be the concentration of NO at equilibrium if 2.00 moles of nitrogen and 6.00 moles oxygen are allowed to come to equilibrium in a 2.0 L flask?
A
1.1 × 10−3 M
B
3.1 × 10−7 M
C
1.2 × 10−6 M
D
6.2 × 10−7 M
Verified step by step guidance1
First, write the balanced chemical equation for the reaction: N₂(g) + O₂(g) ⇄ 2 NO(g).
Calculate the initial concentrations of N₂ and O₂. Since you have 2.00 moles of N₂ and 6.00 moles of O₂ in a 2.0 L flask, use the formula: concentration = moles/volume. For N₂, it is 2.00 moles / 2.0 L = 1.00 M, and for O₂, it is 6.00 moles / 2.0 L = 3.00 M.
Set up an ICE (Initial, Change, Equilibrium) table to track the changes in concentrations. Initially, [NO] = 0 M, [N₂] = 1.00 M, and [O₂] = 3.00 M. Let the change in concentration of NO be +2x, then the change for N₂ and O₂ will be -x each.
Write the expression for the equilibrium constant Kc: Kc = ([NO]²) / ([N₂][O₂]). Substitute the equilibrium concentrations from the ICE table into this expression: Kc = ((2x)²) / ((1.00 - x)(3.00 - x)).
Solve the equation for x using the given Kc value of 4.18 x 10⁻⁷. This will involve simplifying the expression and solving the quadratic equation. The concentration of NO at equilibrium will be 2x.
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