Homework 5I.23
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Homework 5I.23
For number 5I.23 in the homework, why is the initial concentration (in moles) for CH4(g) 0 instead of 0.478?
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Re: Homework 5I.23
Also, would you be able to divide the concentrations (in moles) by 10.0 L in the beginning so the concentrations in the equilibrium table are in mol.L^-1 instead of mol and calculate the concentrations of the different products and reactants from there instead of dividing the moles by 10.0 L at the end after filling out the equilibrium table? Hopefully this question makes sense.
Re: Homework 5I.23
My best guess is that if they don't give you an initial concentration, then it's implied that it's 0. But, I'm not sure what the actual reason is. Sorry, I can't give you a definitive answer but hopefully someone else will have the answer, and you can ask a TA or Lavelle.
And if you get the right answer, I don't think it matters when you divide by the amount of liters. Personally, I think it's easier to do it after filling out the table to divide the moles by the liters.
And if you get the right answer, I don't think it matters when you divide by the amount of liters. Personally, I think it's easier to do it after filling out the table to divide the moles by the liters.
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Re: Homework 5I.23
Eileen Si 3G wrote:For number 5I.23 in the homework, why is the initial concentration (in moles) for CH4(g) 0 instead of 0.478?
The initial concentration of CH4 is zero because we started off with ONLY moles of CO and H2. The 0.478 moles of CH4 is after you wait long enough for the system to reach equilibrium. Therefore, the 0.478 mol CH4 divided by the 10.0L gives you the final, equilibrium concentration of CH4. What you have to do is calculate the change "x" in order to solve for the other concentrations at equilibrium to solve for K. An ICE chart can help you organize your thought process.
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