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.

Video transcript

Hello everyone. Today, we got the following problem. The equilibrium constant for the following decomposition reaction at 75°C is 2.73. What are the equilibrium concentrations of ammonia and boron try fluoride if a sample of this complex is put inside a closed container at 75°C and allowed to reach equilibrium. So one thing to note when writing our equilibrium expressions is that we are only going to include so we could only include our Aquarius and our gasses, meaning solids and liquids are not included with that. In mind. We see that we do have a solid for the reactant. So we're not going to include that. So when we are writing our expression are equally room expression, it's generally the concentration of products over the concentration of reactant. However, we only have gasses for our products. So our expression is going to be the concentration of ammonia times our concentration of boron try fluoride. So based on geometric principles, we can let x which is a variable equal the concentration of ammonia, which can also equal the concentration of boron try fluoride since there's one mole of each and with that we have our case, the expression which can be represented by X times X or X squared R K C expression. According to the question was 2.73. We're gonna equal that two X squared. And when we square both sides, we're going to get 1.65 moller as our final answer. And so with that, we've answered the question overall, I hope this helped. And until next time.