Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    1
    ReputationRep:
    "The standard molar enthalpy of combustion of propanoic acid is -1527.2 kJ mol-1. Given that the standard molar enthalpy change of formation of water is -285.5 kJ mol-1 and that of carbon dioxide is -393.5 kJ mol-1, construct a Hess's Law cycle and calculate the standard molar enthalpy change of formation of propanoic acid?"

    Seriously there are so many ways and different routes people have answered it and its really confused me
    Is there anyone who can give me the one correct equation, correct pathway and correct equation to work this out?
    Offline

    16
    ReputationRep:
    Remember; enthalpy change is a state functions it is not path dependent.

    So, the formation of the propanoic acid followed by the combustion of the propanoic acid, is the same as going straight to the products of the combustion (namely carbon dioxide and water).

    Look the equation of the combustion, work out molar ratios.
    Then calculate the answer.
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.