The reaction

2 PH31g2 + As21g2 ∆ 2 AsH31g2 + P21g2

has Kp = 2.9 * 10-5 at 873 K. At the same temperature,

what is Kp for each of the following reactions?

(a) 2 AsH31g2 + P21g2 ∆ 2 PH31g2 + As21g2

Video transcript

Welcome back everyone. The reaction to so three plus N two in equilibrium with two, no three plus S two has KP of 4.5 multiplied by 10, the power of negative seventh at 1000. Kelvin. What is the KP of the reaction below at the same temperature? Let's analyze the original equilibrium. And it tells us that we have two moles of sulfur trioxide 03, they react with one mole of nitrogen and we are producing two moles of no three and one mole of s too, let's call the equilibrium constant of this reaction. KP one. Our target equation, would it be two, no three plus S two in equilibrium with CSO three plus and two. Let's call the equilibrium constant as KKP two. We want to understand the relationship between these equations. And what we notice is that the reactants of reaction number one become our products and the products of reaction number one become our reactants in reaction. Number two. So we're essentially flipping reaction number one, if we are flipping reaction number one to obtain reaction number two, this is equivalent to multiplying reaction number one by a negative one. This is how we are obtaining reaction number two. Simply speaking, we're taking the reciprocal value of the equilibrium constant to get the second equilibrium constant. So whenever we are reversing our equilibrium, the new equilibrium constant KP two is simply the original equilibrium constant KP one raised to the power of negative one or the reciprocal valley of KP one. So we simply want to take the reciprocal value of 4.5 multiplied by sense, the power of negative seventh, which gives us 2.2 multiplied by 10, the power of sixth. And that's the equilibrium constant of the reverse reaction. Thank you for watching.