Answer :
Answer:
The equilibrium constant Kp for this reaction is 0.0030
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
Answer:
Kp is 3.03 x 10⁻³ to two significant digits
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
[tex]Kp = \frac{|P|C^c * |P|D^d}{|P|A^a * |P|B^b}[/tex]
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
[tex]Kp = \frac{40.7^{2}*65.1^{2}}{76.9^{4}*66.3^{1}}[/tex]
[tex]Kp = \frac{1,656.49 * 4,238.01}{34,970,783.2321 * 66.3}[/tex]
[tex]Kp = \frac{7,020,221.1849}{2,318,562,928.28823}[/tex]
Kp = 0.0030278
Kp = 3.0278 x 10⁻³
Kp is 3.03 x 10⁻³ to two significant digits