The enthalpy change for the reaction of 50.0ml of ethylene with 50.0ml of H2 at

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You are trying to find out the enthalpy (heat) of formation for the burning of Mg which is hard to do all by itself! Instead, if you know the heats of reaction for three other reactions, you can cleverly combine them to come up with the same reaction as the target Mg + 1/2 O2 ---> MgO Here is how! Reaction 1 Mg + 2HCl ---> H2 +MgCl2 H1 = whatever you found from your lab Now take reaction 2 and REVERSE it MgCl2 + H2O ----> MgO + 2HCl H2 = whatever you found from your lab but remember to change the SIGN because you reversed the reaction! If you 'add' these two reactions, see what happens? MgCl2 and 2 HCl drop out! Now add reaction (3) H2 and H20 fall out! Combine the heats of reactions for all three, remembering to reverse the sign for equation (2) and you will have the answer for your target reaction, since it is identical. Cool, eh? Well, really hot, as in exothermic, since burning Mg gives off a lot of heat; now is delta H for this (+) or (-)?

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