Consider the gas phase reaction 2A(g) + 3B(g) ⇌ 2C(g) at a given temperature. When 2.0 moles of A(g) are reacted with 2.0 moles of B(g), 0.8 moles of C(g) are formed at equilibrium at a total pressure of 2.0 bar. The value of the equilibrium constant, KP of this reaction at the given temperature is closest to
2.4
Understanding the Chemical Equilibrium Problem
The question asks us to find the value of the equilibrium constant, \(K_P\), for a given gas phase reaction at a specific temperature. We are provided with the balanced chemical equation, the initial amounts of reactants, the amount of product formed at equilibrium, and the total pressure at equilibrium. To solve this, we need to determine the partial pressures of all gases at equilibrium and then use the \(K_P\) expression.
The given reaction is:
\[ 2A(g) + 3B(g) \rightleftharpoons 2C(g) \]Initial moles:
- A: 2.0 mol
- B: 2.0 mol
- C: 0 mol
Equilibrium moles of C: 0.8 mol
Total pressure at equilibrium: \(P_{total} = 2.0\) bar
Setting up the ICE Table
We can use an ICE (Initial, Change, Equilibrium) table to track the moles of each species in the reaction. Let's denote the change in moles of C as \(+x\). From the stoichiometry of the reaction \(2A(g) + 3B(g) \rightleftharpoons 2C(g)\), if \(x\) moles of C are formed, then \(\frac{2}{2}x = x\) moles of A react, and \(\frac{3}{2}x = 1.5x\) moles of B react.
The equilibrium moles are given by the initial moles plus the change in moles.
| Species | Initial (mol) | Change (mol) | Equilibrium (mol) |
|---|---|---|---|
| A | 2.0 | \( -x \) | \( 2.0 - x \) |
| B | 2.0 | \( -1.5x \) | \( 2.0 - 1.5x \) |
| C | 0 | \( +x \) | \( x \) |
We are given that the moles of C at equilibrium is 0.8 mol. So, \(x = 0.8\).
Now we can find the equilibrium moles of A and B:
- Equilibrium moles of A = \(2.0 - x = 2.0 - 0.8 = 1.2\) mol
- Equilibrium moles of B = \(2.0 - 1.5x = 2.0 - 1.5(0.8) = 2.0 - 1.2 = 0.8\) mol
- Equilibrium moles of C = \(x = 0.8\) mol
Calculating Total Moles and Mole Fractions
The total number of moles at equilibrium is the sum of the moles of all species present at equilibrium.
Total moles, \(n_{total} = n_A + n_B + n_C = 1.2 + 0.8 + 0.8 = 2.8\) mol.
The mole fraction (\(X_i\)) of each gas at equilibrium is given by the moles of the gas divided by the total moles.
- Mole fraction of A, \(X_A = \frac{n_A}{n_{total}} = \frac{1.2}{2.8}\)
- Mole fraction of B, \(X_B = \frac{n_B}{n_{total}} = \frac{0.8}{2.8}\)
- Mole fraction of C, \(X_C = \frac{n_C}{n_{total}} = \frac{0.8}{2.8}\)
Determining Partial Pressures at Equilibrium
According to Dalton's Law of Partial Pressures, the partial pressure of a gas (\(P_i\)) in a mixture is equal to its mole fraction (\(X_i\)) multiplied by the total pressure (\(P_{total}\)).
\(P_i = X_i \times P_{total}\)
Given \(P_{total} = 2.0\) bar.
- Partial pressure of A, \(P_A = X_A \times P_{total} = \frac{1.2}{2.8} \times 2.0\) bar
- Partial pressure of B, \(P_B = X_B \times P_{total} = \frac{0.8}{2.8} \times 2.0\) bar
- Partial pressure of C, \(P_C = X_C \times P_{total} = \frac{0.8}{2.8} \times 2.0\) bar
Let's simplify the fractions before calculating the partial pressures:
- \( \frac{1.2}{2.8} = \frac{12}{28} = \frac{3}{7} \)
- \( \frac{0.8}{2.8} = \frac{8}{28} = \frac{2}{7} \)
So, the partial pressures are:
- \( P_A = \frac{3}{7} \times 2.0 = \frac{6}{7} \) bar
- \( P_B = \frac{2}{7} \times 2.0 = \frac{4}{7} \) bar
- \( P_C = \frac{2}{7} \times 2.0 = \frac{4}{7} \) bar
Calculating the Equilibrium Constant Kp
For the reaction \(2A(g) + 3B(g) \rightleftharpoons 2C(g)\), the expression for the equilibrium constant \(K_P\) is:
\[ K_P = \frac{(P_C)^2}{(P_A)^2 (P_B)^3} \]Substitute the calculated partial pressures into the expression:
\[ K_P = \frac{\left(\frac{4}{7}\right)^2}{\left(\frac{6}{7}\right)^2 \left(\frac{4}{7}\right)^3} \] \[ K_P = \frac{\frac{16}{49}}{\frac{36}{49} \times \frac{64}{343}} \] \[ K_P = \frac{\frac{16}{49}}{\frac{36 \times 64}{49 \times 343}} \] \[ K_P = \frac{16}{49} \times \frac{49 \times 343}{36 \times 64} \]Cancel out the 49 in the numerator and denominator:
\[ K_P = \frac{16 \times 343}{36 \times 64} \]Simplify the fraction:
\[ K_P = \frac{16 \times 343}{2304} \] \[ K_P = \frac{5488}{2304} \]Now, perform the division:
\[ K_P \approx 2.3826 \]The value of \(K_P\) is approximately 2.38. Looking at the given options, the value closest to 2.38 is 2.4.
Final Answer Summary
Based on our calculations, the equilibrium constant \(K_P\) for the reaction at the given temperature is approximately 2.4.
| Parameter | Value |
|---|---|
| Equilibrium moles of A | 1.2 mol |
| Equilibrium moles of B | 0.8 mol |
| Equilibrium moles of C | 0.8 mol |
| Total moles at equilibrium | 2.8 mol |
| Partial pressure of A | 6/7 bar |
| Partial pressure of B | 4/7 bar |
| Partial pressure of C | 4/7 bar |
| Equilibrium constant Kp | \( \approx 2.4 \) |
Important Questions from Chemical Equilibrium
The vapour pressures of water, acetone and ethanol at 293 K are 2.34 kPa, 12.36 kPa and 5.85 kPa respectively. Which of the following has the lowest boiling point?
The equilibrium dissociation energy of a diatomic molecule is 4.75 eV and its stretching frequency corresponds to 0.5 eV. The minimum energy required to dissociate the molecule in eV is
For the reaction N2(g) + 3H2(g) → 2NH3(g), the thermodynamic quantity which depends upon the pressure at which equilibrium is arrived at is (superscript 'o'represents standard state)
The equilibrium constant for the formation of the complex \(\rm Zn (NH_3)_4^{2+}\) at 25°C is close to:
[Given: \(\rm E^0_{Zn^{+2}/Zn}=-0.763V\) and \(\rm E^0_{Zn(NH_3)^{+2}_4/Zn,NH_3}=-1.03V\)]
In the reaction PCl5(g) ⇌ PCl3(g) + Cl2(g), the equilibrium concentration of PCl5 and PCl3 are 0.4 and 0.2 mole/litre respectively. If the value of Kc is 0.5, what is the concentration of Cl2 in mole/litre?
