Could use some help on the following question. Last chemistry class we went quickly over Hess's law but we didn't really try any questions so I could use an example.
Caculate he standard enthalpy change in the reaction
PbO + CO = Pb + CO2
Given the standard enthalpies of formation of lead oxde, carbon monoxide and carbon dioxide are -219, -111 and -394 KJmol^-1 respectively.
Could use some help on the following question. Last chemistry class we went quickly over Hess's law but we didn't really try any questions so I could use an example.
Caculate he standard enthalpy change in the reaction
PbO + CO = Pb + CO2
Given the standard enthalpies of formation of lead oxde, carbon monoxide and carbon dioxide are -219, -111 and -394 KJmol^-1 respectively.
Hess law works because enthalpy change is independent of the route(path) which the reaction takes to happen.
Given standard enthalpy of formation, enthalpy of reaction = total enthalpy of formation of (products - reactants)
Remember to multiply out by no.of mol as shown by your balanced equation.
In your case, all mole ratios are 1, so it is fine, it would be, -394 - (-219 -111) = ?
Okay thankyou I see how that works your working out but what about this question.
Caculate the standard enthalpy of formation of propan-1-onal, CH3CH2CH20H given the standard enthalpies of combustion in KJmol^-1 are -2010 propan-1-ol, carbon -394 and hydrogen -286.
I imagine I would write the reaction so that one mole of propan-1-ol would be fomed. Would I then subtract the the number of carbon and hydrogens from -2010? Also what does the -2010 represent. its not combusting so it isn't realeasing any energy. Is that just the amont of energy it would give out and the energy level its at?
Okay thankyou I see how that works your working out but what about this question.
Caculate the standard enthalpy of formation of propan-1-onal, CH3CH2CH20H given the standard enthalpies of combustion in KJmol^-1 are -2010 propan-1-ol, carbon -394 and hydrogen -286.
I imagine I would write the reaction so that one mole of propan-1-ol would be fomed. Would I then subtract the the number of carbon and hydrogens from -2010? Also what does the -2010 represent. its not combusting so it isn't realeasing any energy. Is that just the amont of energy it would give out and the energy level its at?
Caution : Balanced chemical equation with combustion will most often involve stoichiometric ratios higher than 1, so it is very difficult to just do the calculation without mistake(not that I am saying it is impossible)
Best approach(my opinion) : Draw up a Hess cycle for this. Write an eqn for formation of propan-1-ol, then down from the equation, come up with the balanced eqn for combustion. It shouldn't matter if you have 1 or n mol of propan-1-ol, but you just need to remember to divide your enthalpy change of combustion by this number of mol of propan-1-ol you have in your eqn, ie n mol; but if it is 1, then obviously it isn't going to make a difference.
I wouldn't do the calculation here for you, but you can try it your self, it will be very clear once you get the cycle right. One clue, enthalpy of combustion will always be exothermic, ie negative sign.
Caution : Balanced chemical equation with combustion will most often involve stoichiometric ratios higher than 1, so it is very difficult to just do the calculation without mistake(not that I am saying it is impossible)
Best approach(my opinion) : Draw up a Hess cycle for this. Write an eqn for formation of propan-1-ol, then down from the equation, come up with the balanced eqn for combustion. It shouldn't matter if you have 1 or n mol of propan-1-ol, but you just need to remember to divide your enthalpy change of combustion by this number of mol of propan-1-ol you have in your eqn, ie n mol; but if it is 1, then obviously it isn't going to make a difference.
I wouldn't do the calculation here for you, but you can try it your self, it will be very clear once you get the cycle right. One clue, enthalpy of combustion will always be exothermic, ie negative sign.
Now I'm kind of confused. Just to help clear thins up what does the 2010 for the propan-1-ol represent? Its own energy level?
I pretty sure i am wrong lol but my way was just basically knowing that it's combustion reaction, therefore CH3CH2OH has to react with oxygen to form carbon dioxide and water. Write that out, balance it out which i don't think i've done right tbh. Use the values you are given but in my case since there are 4 carbons, multiply the value by 4 and 5 hydrogens so do the same for that. Then two multiplied by the value for CH3CH2OH and from there just addition i did as shown in my previous post.