Suppose that the gas-phase reactions A → B and B → A are both elementary reactions with rate constants of 4.7×10⁻³ s⁻¹ and 5.8×10⁻¹ s⁻¹, respectively. (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B?

Video transcript

Hello everyone today. We are being given the following problem and asked us to solve for it. So this is a hypothetical gas phase reaction. C to D has a forward rate constant of 4.22 times to the negative second in inverse seconds. The reverse reaction D two C has a rate constant of 1.74 times 10 to the negative fourth. The inverse of seconds. Which component C. R. D has a smaller partial pressure on equilibrium. So the reaction of C. T. D. Or the forward reaction can be denoted as K. C is equal to the constant of the forward reaction divided by the constant of the reverse reaction. And so using this equation we can plug in some numbers here. Our constant for the forward reaction is 4.2, 2 times 10 to the -7 inverse seconds Over the constant for the reverse reaction which is 1. times 10 to the negative 4th in verse seconds. And we plug that in, we get a value of 242.5. Now note that this is a gas phase reaction With the number of moles in equaling zero. Total. We have our K. P. Or our Constant for the pressure of 2 4, 2.5. So there are some rules in relation to the value of K. For example, when K is less than one, there are more reactant than products at equilibrium. When K equals one, the number of reactant and products are equal at equilibrium and when the value of K is greater than one, the there are going to be more products than reactive at equilibrium. And so we see here that our value of KP is 242, which is way greater than one. Therefore there must be more product then reacting at equilibrium. So I'm gonna see more product than the reactant at equilibrium. This would denote that c our reactant has a smaller partial pressure at equilibrium, since there will be more product form or deformed. So with that we have solved our problem. I hope this helped, and until next time.