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Ch.15 - Chemical Equilibrium

Chapter 15, Problem 10a

The diagram shown here represents the equilibrium state for the reaction A2(𝑔) + 2B(𝑔) β‡Œ 2AB(𝑔). (a) Assuming the volume is 2 L, calculate the equilibrium constant 𝐾𝑐 for the reaction.

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Hello everyone today. We have the following problem. The diagram below shows nitric oxide reacting with oxygen to form nitrogen dioxide. And we have the following equation here. Using the information in the diagram calculate the value of the equilibrium constant K C. For the reaction, assume the volume of the container is 3.4 million or 3. liters. So we have this equation here and we know that our Casey equilibrium is going to be equal to the products or the concentration of the products over the concentration of the reactant. And so we fill this out for our products. We just have our nitrogen dioxide or N. 02 and that coefficient two is going to become our exponent. And so far reactant we have nitrous nitrous oxide and that coefficient of two will become an exponent. And for our last reactant we have just oxygen gas with no coefficient, so no exponents. And so next after that we have we have to calculate the polarity for each component. So for our nitric oxide, malaria is equal to moles over L. So we know that we have four moles of our, you know, nitric oxide is going to be one red and one blue. So we have one red, one blue here, we have another one red, one blue, we have another one red, one blue and we have one final one red and one blue and that's going to constitute four different molds. And so we have four moles over the leaders which is our 3.4 liters that we were given in the question stem. And this gives us about 1.177 polarity for our oxygen. That's gonna be to read. So we're gonna have 12 right there, 2345 and six. So therefore we have six moles of oxygen gas divided by that 3. liters. To give us 1.76 polarity. Last but not least we have our nitrogen dioxide which is going to be two oxygen's or two red circles in one blue. So therefore we have two moles of those Or I misspoke there, we have three moles of those since we have one final one down there. So we have three moles of that over this 3.4 L. And that's gonna give us about 0.88 malaria. And so using that we can plug that into this equation here, That's gonna be a 3rd step there. So we have the concentration of R. N. 02 or nitrate nitrogen dioxide, which is 0.88. We're going to square that. We're going to put that over the concentration of our nitrogen nitric oxide, which is 1.177 malaria. And we're also gonna square that. And then we're gonna multiply that finally by the concentration of oxygen gas which is 1.76. And we do that. We get the value of K. C. is equal to about 0.32. And with that we have found our answer. I hope this helped. And until next time.
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