At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium. 2SO2(g) + O2(g) ? 2SO3 (g) What is Kc at this temperature ?

To find the equilibrium constant \( K_c \) for the reaction given \( K_p \) at a specific temperature, we can use the relationship between \( K_p \) and \( K_c \). The equation that connects these two constants is:

Understanding the Relationship

The relationship between \( K_p \) and \( K_c \) is defined by the equation:

K_p = K_c (RT)^{\Delta n}

In this equation:

• K_p is the equilibrium constant in terms of partial pressures.
• K_c is the equilibrium constant in terms of molar concentrations.
• R is the ideal gas constant, which is approximately 0.0821 L·atm/(K·mol).
• T is the temperature in Kelvin.
• Δn is the change in the number of moles of gas, calculated as the moles of gaseous products minus the moles of gaseous reactants.

Calculating Δn

For the reaction:

2SO₂(g) + O₂(g) ⇌ 2SO₃(g)

We can determine \( Δn \) as follows:

• Products: 2 moles of SO₃
• Reactants: 2 moles of SO₂ + 1 mole of O₂ = 3 moles

Thus, \( Δn = 2 - 3 = -1 \).

Substituting Values

Now, we can substitute the values into the equation:

K_p = K_c (RT)^{\Delta n}

Given:

• At 450 K, \( K_p = 2.0 \times 10^{10} \, \text{bar} \)
• R = 0.0821 L·atm/(K·mol) (we will convert this to bar for consistency)

To convert R to bar, we can use the fact that 1 atm = 1.01325 bar, so:

R = 0.0821 L·(1.01325 bar)/(K·mol) ≈ 0.08314 L·bar/(K·mol)

Now, substituting the values into the equation:

2.0 × 10^{10} = K_c (0.08314 × 450)^{-1}

Calculating \( RT \):

RT = 0.08314 × 450 ≈ 37.413

Now, we can rewrite the equation:

2.0 × 10^{10} = K_c (37.413)^{-1}

Calculating \( (37.413)^{-1} \):

(37.413)^{-1} ≈ 0.0267

Now, substituting back into the equation gives:

2.0 × 10^{10} = K_c × 0.0267

Solving for Kc

To find \( K_c \), we rearrange the equation:

K_c = \frac{2.0 × 10^{10}}{0.0267}

Calculating this gives:

K_c ≈ 7.48 × 10^{11} \, \text{mol/L}

Final Result

Therefore, the equilibrium constant \( K_c \) at 450 K for the reaction is approximately:

K_c ≈ 7.48 × 10^{11} \, \text{mol/L}