Calculate the value of the equilibrium constant at 427 °C for the reaction

Na O1s2 + 1>2 O 1g2 ∆ Na O 1s2

given the following equilibrium constants at 427 °C.

Na2O1s2 ∆ 2 Na1l2 + 1>2 O21g2 Kc = 2 * 10-25 Na O 1s2 ∆ 2 Na1l2 + O 1g2 K = 5 * 10-29

Video transcript

Welcome back, everyone. Given the following data at 1500 Kelvin calculated the equilibrium constant at the same temperature for the reaction to fe 304 plus one half 02 in equilibrium with three fe 203. Let's look at the given reactions to us. We understand that they have species involved in the target equation. So essentially, we want to manipulate reactions one and two in such a way that we obtain the target equation somehow. And one of the easiest ways is to analyze each species in the target equation. We're going to begin with our first reactant which is fe 304 and locate it in the second equation given to us. However, we understand that there's a coefficient of two in the target equation and we have a coefficient of one in equation number two, meaning we want to multiply the whole reaction number two by two to get that coefficient of two. So let's go ahead and do that. We obtain two fe three or four in equilibrium, we now three becomes six because we're multiplying by two. So six fe plus 4 +02. Now what happens to the equilibrium constant K two. If we are multiplying by two, we need to raise K two to the power of two because the multiplier becomes the exponent of the equilibrium constant, which allows us to get the new equilibrium constant for the modified equilibrium well done. Now, we want to think about fe 203. And the reason why we're skipping oxygen is because we have oxygen in both of our equations. So we can easily locate fe 23 in the first equilibrium. But now we have the coefficient of three in the target equation, meaning we want to multiply the whole equilibrium number one by three. And we also want to reverse it because fe 23 is on the product side for the target equation and it is on the reactant side or the given equilibrium number one, reversing essentially means multiplying by negative one. So if we combine negative one and three, the net result would be multiplication by negative three. What do we get? Well, let's write down the resultant equilibrium, we will have two multiplied by three, which gives us six fe plus, we have multiplied by three, gives us nine apps 02. In equilibrium words, we fe 23, we can already tell the expression of the equilibrium constant. For this one. It will be K one raised by negative three because that's our multiplier well done. Now we want to simply add them up. The reason why we're adding them up is because we already have our reactants and products at those places where we need them to be. Right. And what we notice immediately is that we have six moles of iron on opposite sides of the equilibrium arrows. So we can just cancel them out. And this gives us two fe 304. What about oxygens? We have nine halves and four on the right hand side, four is the same as 8/2. So 9/2 minus 8/2 gives us one half oxygen. On the left hand side, which is exactly what we wanted in equilibrium West, we fe 203, we have successfully checked if the resultant equation gives us the target equation. And we have shown that yes, indeed, we have managed to get the target equation. And since we added them up, this means multiplying the individual equilibrium constants. So the equilibrium constant KC is simply K two squared multiplied by key one to the power of negative third or simply K two squared divided by K one Q. And we just want to substitute the given values starting with K two. We're going to take 7.0 multiplied by 10. The power of negative 22nd squared and divide by K one cubed which is 5.0 multiplied by 10 to the power of negative 18th heaved. If we evaluate the result, we get 3.9 multiplied by sense. The power of ninth that's the equilibrium constant for the target equilibrium. Thank you for watching.