Hi,
Can someone explain how to calculate Hrxn? Is there an equation that we should be using?
Achieve 7 (Week 5)
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Re: Achieve 7 (Week 5)
Hi! For this question, you're going to need to look at the Hf values for each molecule (the first column). For instance, CO2(g) is –393.5 kJ/mol. Continue finding the rest of the enthalpy values, and then multiply each enthalpy by the coefficients present in the reaction. For example, if CO2 has 3 mols, you would multiply –393.5 x 3. Afterward, you will subtract products - reactants to reach the final enthalpy change.
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Re: Achieve 7 (Week 5)
The equation you use is Delta H = sum of Delta H formation (products) - (sum of Delta H formation(reactants)). you can find the delta H of formation for each individual product and reactant on the table under Delta Hf.
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Re: Achieve 7 (Week 5)
∆Hrxn=∆Hf(products)-∆H(reactants) is the equation you'd use, in this instance. Hope this helps!!
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Re: Achieve 7 (Week 5)
Because you are trying to find the enthalpy change = delta H of reaction = Delta H of product - Delta H of reaction. To find Delta H of the product you multiply the standard enthalpy values for each chemical (found on the list that is given in the problem - just click on the link) by the amount of it (as seen in the stoichiometry - # of moles each has). It is the same thing for Delta H of reaction but obviously just looking at the reaction side. You subtract the two to find the amount of enthalpy change.
Keep in mind to label everything/use units so that you can cross it out along the way so you can make it clear to yourself why the final units should be kJ.
Keep in mind to label everything/use units so that you can cross it out along the way so you can make it clear to yourself why the final units should be kJ.
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Re: Achieve 7 (Week 5)
Hi! For this problem, you should use enthalpy of formation values to find deltaHrxn. Use the formula Delta H = sum of Delta H formation (products) - sum of Delta H formation(reactants) using the standard enthalpy of formation values for reach reactant and product. Hope that helps!
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