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Ch.5 - Thermochemistry

Chapter 5, Problem 70a

Write balanced equations that describe the formation of the following compounds from elements in their standard states, and then look up the standard enthalpy of formation for each substance in Appendix C: (a) CH3OH(l)

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Hi everyone for this problem. It reads identify the elements in their standard states required for the formation equation of glucose. And write the balanced reaction. Find the standard entropy of the formation using the standard table of values. Okay, so the first thing that we want to do is we want to identify the elements of glucose in their standard state and then we're going to write a balanced equation and then we're going to find the standard entropy of formation. Alright, so for glucose we need to figure out the elements in their standard state. Okay, so glucose is written as C six H 12 06. Okay, so the elements and glucose our carbon, hydrogen and oxygen. But we need to write these out in their standard states. Okay, so carbon and its standard state is going to be solid carbon. Okay, hydrogen and its standard state is going to be hydrogen gas and oxygen and its standard state is going to be oxygen gas. Okay, so this is going to be the first part to our problem, the elements in their standard states. Okay, the second part of the question asks us to write the balanced reaction. Okay, so we know we have glucose as our final product. Okay, so we need to figure out what is our balanced reaction of our elements that's going to create this product. Okay, so we'll go ahead and write that out next. So we have glucose and we need to figure out what are our reactant since. Okay, so we know we need carbon. Okay, looking at this, we know we need six carbons and so we can start off with six carbons in its standard state. Okay, the next thing we know is that we need 12 hydrogen. Okay, but we know that hydrogen in its standard state is H two gas. Okay, so to balance this out, that means we're going to need six hydrogen gasses in its standard state. Okay. And lastly, we have six oxygen's okay. So we know that oxygen in its standard state is owed to gas. So what we need to make six of this is we need to put a three here. Okay, so that three is going to have that six oxygen gas in the product side. Okay, so this is going to be our balanced reaction. Alright, and the last part of the question is the standard entropy of formation. This is a value that we can look up. And that value for the standard heat of formation for glucose is negative 1,273.0 to kill a jules per mole. Okay, so there it is. That is the last part to this three part question. And that is it for this problem. I hope this was helpful