Question

If the equilibrium constant kc for the reaction below is 1.71 × 10–1, what will be the equilibrium pressure of no if the initial partial pressures of the three gases are all 1.96 × 10–3 atm? $$

Answer 1

To determine the equilibrium pressure of NO, set up an ICE table using the values of the equilibrium constant (Kc) and initial partial pressures. Assume the increase in NO pressure is 'x' and the decrease in N2O and O2 pressures is '2x' and 'x', respectively. Calculate the equilibrium pressures by adding the initial pressures to their corresponding changes.

Step-by-step explanation:

To determine the equilibrium pressure of NO, we need to set up an ICE table and use the values of the equilibrium constant (Kc) and initial partial pressures of the three gases. Let's assume the increase in the pressure of NO is 'x'. Since the stoichiometric coefficient of NO is 2, the decrease in the pressures of N2O and O2 will be '2x' and 'x', respectively. The equilibrium partial pressures of the three gases can be calculated by adding the initial pressures to their corresponding changes.

Initial pressures:

• N2O: 1.96 × 10-3 atm
• O2: 1.96 × 10-3 atm
• NO: 1.96 × 10-3 atm

Change in pressures:

• N2O: -2x
• O2: -x
• NO: x

Equilibrium pressures:

• N2O: (1.96 × 10-3 - 2x) atm
• O2: (1.96 × 10-3 - x) atm
• NO: (1.96 × 10-3 + x) atm

Since the equilibrium constant (Kc) is given as 1.71 × 10-1, which represents the ratio of the equilibrium concentrations of products to reactants, we can set up the following equation:

Kc = [NO]2 / ([N2O] * [O2])

Substituting the equilibrium pressures into this equation:

1.71 × 10-1 = ([1.96 × 10-3 + x]2 / ([1.96 × 10-3 - 2x] * [1.96 × 10-3 - x]))

Now, solve this equation to find the value of 'x', which represents the equilibrium increase in the pressure of NO.

Once 'x' is determined, substitute it back into the equations for the equilibrium pressures to calculate the final equilibrium pressures of N2O, O2, and NO.

Answer 2

The equilibrium reaction that is not shown is the reaction of N₂ and O₂ to form NO:

N₂ + O₂ ⇄ 2NO

We can write an expression for the equilibrium constant. We are given the value of Kc, but we must use Kp if we are dealing with partial pressures. However, since 2 moles of gas are formed from 2 moles of gas, the value of Kc is equal to Kp so we do not need to change anything.

Kc = (Pno)²/(Pn2)(Po2) = 1.71 x 10⁻¹

We must first use the initial pressures to find the reaction quotient to decide which way the equilibrium will shift:

Q = (1.96 x 10⁻³)²/(1.96 x 10⁻³)(1.96 x 10⁻³) = 1

Since Q > Kp, the equilibrium will shift to the left. Now we can prepare an ICE table to find the equilibrium pressures:

N₂ O₂ NO
I 1.96 x 10⁻³ 1.96 x 10⁻³ 1.96 x 10⁻³
C +x +x -2x
E 0.00196 + x 0.00196 + x 0.00196 - 2x

We enter these equilibrium pressures into the equilibrium expression and solve for x:

0.171 = (0.00196 - 2x)²/(0.00196 + x)²

Expand the equation and simplify to a quadratic function:

3.829x² - 0.00851x + 3.183·10⁻⁶ = 0

x = ((0.00851) +/- sqrt((0.00851)² - 4(3.829)(3.183 x 10⁻⁶)))/2(3.829)

x = 0.000476 atm

Now we can solve for the equilibrium partial pressures:

Pno = 0.00196 - 2(0.000476) = 0.00101 atm

Pn2 = 0.00196 + 0.000476 = 0.00244 atm

Po2 = 0.000196 + 0.000476 = 0.00244 atm

Therefore, the equilibrium partial pressure of NO is 0.00101 atm.