At 80°C, 𝐾_𝑐= 1.87×10⁻³ for the reaction PH₃BCl₃(𝑠) ⇌ PH₃(𝑔) + BCl₃(𝑔) (a) Calculate the equilibrium concentrations of PH₃ and BCl₃ if a solid sample of PH₃BCl₃ is placed in a closed vessel at 80°C and decomposes until equilibrium is reached. (b) If the flask has a volume of 0.250 L, what is the minimum mass of PH₃BCl₃(𝑠) that must be added to the flask to achieve equilibrium?

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Hey, everyone, we're told that the equilibrium constant for the following decomposition reaction at degrees Celsius is 2.73. What is the minimum amount of ammonia? Try flora boring Ingram's that has to be added to a 0.3 50 liter flask. In order to reach equilibrium. First, let's go ahead and write out our equilibrium expression and use that to determine the concentrations of ammonia and boron try fluoride for this specific reaction. We know that our equilibrium expression is going to be equal to the concentration of our ammonia times the concentration of our boron try fluoride. Now, the reason we do not divide this by the concentration of ammonia trifle or a boring is because it is a solid and we do not include solids and liquids in our equilibrium expression. So let's go ahead and let X equal the concentration of our ammonia, which also means it is the concentration of our boron try fluoride. Since based on so geometry, the concentrations are going to be equal. So we have our equilibrium expression and it's going to be equal to X times X. Now, when we plug in these values, we were told that our equilibrium expression is 2.73. So this is going to be equal to X squared solving for X, we're going to take the square root of both sides. So this means X is going to be equal to 1. moller. So this is our equilibrium concentration. So since the mole ratios are 1-1, this means that the mole of ammonia or the mole of boron try fluoride produce is equal to the mole of ammonia trifle or a boring required. So if we go ahead and take that 0.350 l and use dimensional analysis, we're going to take that 1.65227 moles over leaders of our ammonia and we're going to use that multiple ratio. So based on our reaction, we know that one mole of ammonia is equivalent to one mole of ammonia. Try flora boring. So when we calculate this out and cancel out all of our units, we end up with 0.578 to nine mol of our ammonia. Try flora boring. So let's go ahead and use this value of 0.578- mol to calculate the mass required. So using the molar mass of ammonia, try flora boring, we have 84.84 g of ammonia, try flora boring per one mole of ammonia, try flora boring. So when we calculate this out we end up with 49.0625 g. So simplifying this value, our answer is going to be 49.1 g. Now, I hope this made sense and let us know if you have any questions.

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Key Concepts

Equilibrium Constant (Kc)

Le Chatelier's Principle

Stoichiometry and Concentration Calculations