Delta Fares

Mastering Chemistry: What is the delta H for this reaction?

What is the enthalpy change, delta H degree sign rxn, for this reaction? Express your answers in kilojoules to the tenths place. 2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g) The delta H values I am given are: KOH(aq) = -482.4 kJ/mol H2O(l) = -285.83 kJ/mol I keep getting the answer -393.14, so I have tried the answers -393.1, -393.14, -393.2 and it says that it is not correct. My feedback says "Not quite. Check through your calculations; you may have made a rounding error or used the wrong number of significant figures." So I am completely confused about how it is wrong because other sources have told me that my answer is correct, if anyone is able to help me out with this I would be very appreciative and thankful. Thanks! Thanks IlovsMaths but I tried it and that didn't work either :( got the same feedback too. I have 4 tries left Nope, didn't work m w but thanks for trying to help me

Public Comments

1. Delta H for the reaction = Delta H (products) - Delta H (reactants) = 2 * -482.4 - 2 * -285.83 = -964.8 + 571.66 = -393.14 kJ mol^-1 = -393 kJ mol^-1 (correct to 3 sig. fig.) MasteringChemistry normally looks for answers rounded to three significant figures. Try -393... I hope that helps. :) ILoveMaths07.
2. most definitely... dHrxn = dHproducts - dHreactants = 2x(-482.4) - 2x(-285.83) = -964.8 + 571.67 = -393.1 with 4 sig figs. the reason that has 4 sig figs is that you're adding in the last step. you go by the number with the lowest precision. -964.8 is precise to the 0.1 column. So your number can only be precise to the 0.1. ********** it could be that the value of KOH(aq) should have been. -384.37.. from here... http://www.uwsp.edu/chemistry/tzamis/chem105pdfs/Formation_Enthalpies.pdf in which case... 2x(-482.37) - 2x(-285.83) = -964.74 + 571.67 = -393.07 with 5 sig figs. but I doubt that. it could also be that the question is looking for this... 2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g) + 393.1 kJ as opposed to this... 2K(s) + 2H2O(l) -> 2KOH(aq) + H2(g).... dHrxn = -393.1 kJ so you could try changing the sign.