Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PCl₃(𝑔) + Cl₂(𝑔) ⇌ PCl₅(𝑔). A 7.5-L gas vessel is charged with a mixture of PCl₃(𝑔) and Cl₂(𝑔), which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are 𝑃_PCl3 = 0.124 atm, 𝑃_Cl2 = 0.157 atm, and 𝑃_PCl5 = 1.30 atm. (a) What is the value of 𝐾_𝑝 at this temperature?

Video transcript

hi everyone. So we may have sulfur dioxide gas reacting with oxygen gas to produce sulfur oxide. We have a 5.5 liter reaction vessel that's charged with the mixture of sulfur dioxide and oxygen gas at 300 degrees Celsius. When the equilibrium was established to make sure that the three gasses that these partial pressures for the partial pressure of sulfur dioxide. We have 1.15 atmosphere for the partial pressure of oxygen gas. We have 2.35 atmosphere for the partial pressure of sulfur trioxide. We have 0.855 atmosphere. Who asked to determine the K. P. Just for us to find the K. P. When we're giving the partial pressure of the gas is we need to look at the K. P. Expression for the reaction we call that the K. P. Expression is the partial pressure of the products raised to the power of the coefficient provided by the pressure pressure of the reactant. Right to the power of the coefficient. So for this reaction we have K. P. Because the partial pressure of sulfur trioxide raise to the power of two because we have two moles of sulfur dioxide in the reaction divided by the partial pressure of sulfur dioxide. The power of two because we have two moles of sulfur dioxide and the reaction pressure pressure of 02. For the power of one. For the partial pressure of sulfur trioxide, We have 0.855 atmosphere for the partial pressure of sulfur dioxide. We have 1.15 atmosphere for the partial pressure of oxygen gas, We have 2.35 atmosphere. If we plug in the values for K. P, We're gonna get 0.855 atmosphere where Divide by 1.15 atmosphere square Times 2.35 atmosphere for K. P. We get 0.235 K. P. Does not have any units. Thanks for watching my video, and I hope it was helpful.