Answer:
The equilibrium partial pressure of O2 is 0.545 atm
Explanation:
Step 1: Data given
Partial pressure of SO2 = 0.409 atm
Partial pressure of O2 = 0.601 atm
At equilibrium, the partial pressure of SO2 was 0.297 atm.
Step 2: The balanced equation
2SO2 + O2 ⇆ 2SO3
Step 3: The initial pressure
pSO2 = 0.409 atm
pO2 = 0.601 atm
pSO3 = 0 atm
Step 4: Calculate the pressure at the equilibrium
pSO2 = 0.409 - 2X atm
pO2 = 0.601 - X atm
pSO3 = 2X
pSO2 = 0.409 - 2X atm = 0.297
X = 0.056 atm
pO2 = 0.601 - 0.056 = 0.545 atm
pSO3 = 2*0.056 = 0.112 atm
Step 5: Calculate Kp
Kp = (pSO3)²/((pO2)*(pSO2)²)
Kp = (0.112²) / (0.545 * 0.297²)
Kp = 0.261
The equilibrium partial pressure of O2 is 0.545 atm
Answer:
B. products are found on the write side of the arrow in a chemical reaction.
The
correct answer is A. In the combined gas law, if the volume is decreased and
the pressure is constant, then the temperature decreases.
<span>P1V1/
T1 = P2V2 / T2</span>
<span>Assume
the volume decrease by half; V2 = V1/2</span>
<span>P1V1/
T1 = P2V1 /2 T2</span>
<span>Cancelling
terms,</span>
<span>1/T1
= 1/2 T2</span>
T2
= T1/2
<span>Thus,
the temperature decreased.</span>
Answer:
Where is the chemical equations?
