Nitric oxide (NO) reacts readily with chlorine gas as follows: 2 NO(𝑔) + Cl₂(𝑔) ⇌ 2 NOCl(𝑔) At 700 K, the equilibrium constant Kp for this reaction is 0.26. Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium. (b) P_NO = 0.12 atm, P_Cl2 = 0.10 atm, P_NOCl = 0.050 atm

Video transcript

hi everyone for this problem. It reads the reaction of phosphorous tri chloride with oxygen gas occurs readily. As shown the equilibrium constant. K P for the reaction is 1.2 at 850 kelvin. Will the reaction be at equilibrium when the following atmospheric pressures are mixed? At 850 kelvin? If not, does the mixture need to produce more products or react ints to attain equilibrium? So the question we're trying to answer and this problem is will the reaction be at equilibrium at the following atmospheric pressures? And a key piece of information we're given in this problem is the K. P value, which is our equilibrium pressure. Okay. And what we're going to need to do for this problem is compare K. R equilibrium constant to Q. Which is our non equilibrium constant. Okay. So when K is greater than Q. That means the forward direction is going to be preferred. Okay. It's going to be favored. So we can say the forward direction is favored when K is greater than Q. When K is less than Q. The reverse direction is favored. However, when K is equal to Q, that means we're at equilibrium. So the question is asking does K equal Q? Okay. And if K does equal Q. We're at equilibrium, we know what our K is but we don't know what Q is. So that's what we're going to need to solve for in this problem to compare the two values to see if we're at equilibrium in the problem? We're told R K is equal to 1.0 to our equilibrium expression for K is our partial pressures of our products raised to their stoke eo metric coefficient over the partial pressure of our reactant raised to their spokeo metric coefficients. Okay, so our for Q. This is the value we're looking for. We don't know it, but the expression is going to be the same. So the only thing that's different is our partial pressures that are going into our expression are non equilibrium pressures, which are the pressures that were given in the problem. Okay. So because we know the values we can go ahead and just plug those values in for Q and we'll get the value. Okay, so we're going to get Q is equal to the partial pressure of our product raised to its Tokyo metric coefficient. So, we see in the problem it was given as .25 atmospheric pressure. And this is going to be squared. And the reason it's squared is because it's being raised to its Tokyo metric coefficient. So we have two moles of it. So that means that two is now going to become an exponent over the partial pressure of our reactant. So we have 0.018 atmospheric pressure squared for our phosphorous tri chloride. And then for oxygen gas at 0. atmospheric pressure. Okay. And so that gives us a Q equal to 12,056. So now we know what Q is and we know what K is. So we're going to compare the two. So Kay is 1.02 and Q is 12,056. So here we can see that our K is less than Q. So if we go back up we see that if K is less than Q, that means the reverse direction is going to be favored. Okay. So we can go ahead and say the reverse direction is favored. So to answer the question, will the reaction be at equilibrium? The answer is no, no, it is not at equilibrium because K does not equal cute. Okay. And the second part says if it's not, does the mixture need to produce more product or reactant to obtain equilibrium? Because K is less than Q. This means the reverse direction is favored. If the reverse direction is favored, that means the our reaction is going to shift in the reverse to reach equilibrium. So it's going to produce more reactant. Okay. So we'll say the mixture will need to produce more react mints to attain equilibrium. Okay, so that is the answer to this problem and that is the end of this problem. Okay, I hope this was helpful