You are watching: For which one of the following reactions will δh be approximately (or exactly) equal to δe?
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For the first problem, you"re right in suggesting that, if there is no work done, ∆H is approximately equal to ∆U. Work would be done if, for instance, the pressure changes because more gas is produced or consumed compared to what you started with. So, the answer is (a) because two moles of gas will produce two moles of gas - thus, there is approximately no net change in pressure. The other reactions will have a change in the number of moles of gas present.For the second question: ∆Hºf is the enthalpy of formation for a substance from their respective elements at 25ºC and 1 atm of pressure. ∆H is a more general term in that it is the heat of reaction for any reaction at any temperature or pressure (the "º" restricts the latter clause to 25ºC and 1 atm of pressure, since that"s "standard" thermodynamic conditions). You can determine ∆Hº for a reaction from the ∆Hºf of the respective species in the reaction, but I"m not sure if you"ve covered that yet in your class.
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Thanks for your help. I"m still a little fuzzy on this material, so the other three reactions are not the answer because:b) work is done because the water changes state?c) no moles of gas in the reactant, but one mole of gas in the products?d) same number of moles of reactants as products, but 2 moles of gas producing one mole of gas?So a change of state means a change in pressure?
b) The work being done is the water expanding to become steam and this gas pushing on the surroundings.c) yep. same idea - more gas to push on the surroundings.d) yep - this time, work is done on the system since the volume and/or pressure of the gas is reduced.And liquid gas changes results in a substantial pressure or volume change (depending on the conditions), and generates work either way. Solid liquid has pretty negligible volume changes, though.
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Thank you. The teacher never really went over this, and I couldn"t find the answer in the text book or my notes.