Question

Hydrogen chloride and oxygen react to form chlorine and water, like this:

4 HCl(g) + O2(g) → 2 Cl2(g) + 2 H2O(g)

Also, a chemist finds that at a certain temperature the equilibrium mixture of hydrogen chloride, oxygen, chlorine, and water has the following composition:

COMPOUND Pressure at equilibrium
HCl 76.9 atm
O2 66.3 atm
Cl2 40.7 atm
H2O 65.1 atm

Calculate the value of the equilibrium constant Kp for this reaction. Round your answer to 2 significant digits.

Answer 1

Kp is 3.03 x 10⁻³ to two significant digits

Step-by-step explanation:

Kp is the ratio of partial pressure of gaseous produce raised to power of their stiochiometric to the partial pressure of gaseous reactant raised to power of their stiochiometric.

For a reaction: aA + bB ==> cC + dD

`Kp = (|P|C^c * |P|D^d)/(|P|A^a * |P|B^b)`

Relating the expression with the given expression and data

A = HCl 76.9 atm with a = 4

B = O₂ 66.3 atm with b = 1

C = Cl₂ 40.7 atm with c = 2

D = H₂O 65.1 atm with d = 2

`Kp = (40.7^(2)*65.1^(2))/(76.9^(4)*66.3^(1))`

`Kp = (1,656.49 * 4,238.01)/(34,970,783.2321 * 66.3)`

`Kp = (7,020,221.1849)/(2,318,562,928.28823)`

Kp = 0.0030278

Kp = 3.0278 x 10⁻³

Kp is 3.03 x 10⁻³ to two significant digits

Answer 2

The equilibrium constant Kp for this reaction is 0.0030

Step-by-step explanation:

Step 1: Data given

Partial pressure at the equilibrium:

pHCl = 76.9 atm

pO2 = 66.3 atm

pCl2 = 40.7 atm

pH2O = 65.1 atm

Step 2: The balanced equation

4 HCl(g) + O2(g) → 2 Cl2(g) + 2 H2O(g)

Step 3: Calculate the equilibrium constant Kp

Kp = ((pH2O²)*(pCl2²)) /(pO2)*(pHCl^4))

Kp = (65.1²*40.7²) / (66.3*(76.9^4))

Kp = 0.0030

The equilibrium constant Kp for this reaction is 0.0030