Question

Air (79 percent \(\mathrm{N}_{2}\) and 21 percent \(\mathrm{O}_{2}\) ) is heated to \(2000 \mathrm{K}\) at a constant pressure of 2 atm. Assuming the equilibrium mixture consists of \(\mathrm{N}_{2}, \mathrm{O}_{2},\) and \(\mathrm{NO},\) determine the equilibrium composition at this state. Is it realistic to assume that no monatomic oxygen or nitrogen will be present in the equilibrium mixture? Will the equilibrium composition change if the pressure is doubled at constant temperature?

Short answer

Question: Find the equilibrium composition of a mixture of N2, O2, and NO at 2000 K and 1 atm pressure, and determine if it is realistic to assume that no monatomic oxygen/nitrogen will be present in the mixture. In addition, discuss the changes in the equilibrium composition of the mixture due to changes in pressure. Answer: At 2000 K and 1 atm pressure, the equilibrium partial pressures are approximately N2: 0.7895 atm, O2: 0.2095 atm, and NO: 5.47 x 10^(-4) atm. Due to the low amount of NO, it is realistic to assume that no monatomic oxygen or nitrogen will be present in the mixture. If the pressure is doubled at constant temperature, Le Chatelier's principle predicts that the equilibrium will shift towards the side with fewer moles (reactants), increasing concentrations (partial pressures) of N2 and O2 and decreasing that of NO.

Step-by-step solution

Write the chemical equation.

For the equilibrium composition, the chemical equation is as follows: $$\mathrm{N}_{2} + \mathrm{O}_{2} \rightleftharpoons 2\mathrm{NO}$$

Write the equilibrium expression.

Using the law of mass action, the equilibrium expression is given by: $$K_p = \frac{([\mathrm{NO}]^2)}{([\mathrm{N}_{2}][\mathrm{O}_{2}])}$$ Where \(K_p\) is the equilibrium constant (in terms of partial pressures), [NO], [N2], and [O2] are the partial pressures of NO, N2, and O2 respectively.

Find the equilibrium constant (\(K_p\)).

The equilibrium constant can be found using the given temperature and the relation: $$\log K_p = -\frac{\Delta H^\circ}{2.303R}\times\frac{1}{T} + \frac{\Delta S^\circ}{2.303R}$$ For the reaction at hand, we have \(\Delta H^\circ = -180.3\ \mathrm{kJ/mol}\) and \(\Delta S^\circ = -0.104\ \mathrm{kJ/mol\cdot K}\), T = \(2000\ \mathrm{K}\), and R = \(8.314\ \mathrm{J/mol\cdot K}\). Plugging in these values, we get: $$\log K_p = \frac{180.3 \times 10^3}{2.303 \times 8.314} \times \frac{1}{2000} - \frac{104}{2.303 \times 8.314}$$ Solving this, we find: $$K_p = 2.83 \times 10^{-6}$$

Set up a table with partial pressures.

Let 'x' be the partial pressure of NO in moles. Therefore, the initial partial pressures (mole fractions) are: N2: \((0.79-x)\) atm O2: \((0.21-x)\) atm NO: \(x\) atm

Solve for the equilibrium partial pressures.

Substituting these partial pressures values into the equilibrium expression, we get: $$2.83 \times 10^{-6} = \frac{x^2}{((0.79-x)(0.21-x))}$$ Now, we solve the quadratic equation for x (using the quadratic formula or other methods), and we find two possible values for x: \(x = 5.47 \times 10^{-4}\) or \(x = 0.2105\) From this, we can find the partial pressures and mole fractions of all species in the final mixture.

Determine which x value to consider.

If we choose \(x = 0.2105\), this would mean that \([O_2] = -x \approx -0.21\), which is not possible as a partial pressure cannot be negative. Therefore, we will consider \(x = 5.47 \times 10^{-4}\). The final equilibrium partial pressures are: N2: \((0.79 - 5.47 \times 10^{-4})\) atm O2: \((0.21 - 5.47 \times 10^{-4})\) atm NO: \(5.47 \times 10^{-4}\) atm

Check assumption for monatomic species.

Since the amount of NO in the equilibrium mixture (x) is very small, it is unlikely that there will be any significant amount of monatomic nitrogen or oxygen present in the mixture. Therefore, it is realistic to assume that no monatomic oxygen or nitrogen will be present.

Discuss pressure effect on equilibrium composition.

Changing the pressure at constant temperature does not affect the equilibrium constant (\(K_p\)). However, it affects the mole fractions of species in the reaction mixture. In this case, the equilibrium reaction has 1 mole of reactants and 2 moles of products, so Le Chatelier's principle predicts that increasing the pressure at constant temperature should favor the side with fewer moles (reactants). Therefore, if the pressure is doubled at constant temperature, we would expect the concentrations (and partial pressures) of N2 and O2 to increase and that of NO to decrease.