Achieve Q7 Week 3

[Teresa Dinh 3L]

Hi,

The question reads:

Calculate the standard enthalpy change for the reaction at 25 ∘ C. Standard enthalpy of formation values can be found in this list of thermodynamic properties.

C3H8(g)+5O2(g)⟶3CO2(g)+4H2O(g)

I added -1180.5 (3CO2) and -1143.2 (4H2O) together, got -2323.7 and subtracted it by -103.8 (C3H8) and got -2219.9 but this was wrong - can someone explain why?

[Caroline 2A]

Hi!

It should be calculated [3mol x (-393.509 kJ/mol)] + [4mol x (-241.818 kJ/mol)] - [1mol x (-103.8 kJ/mol)] - [5mol x 0 kJ/mol)]
Which gives you a number close to the number you got but not exactly.

Hope this helps :)

[Rachel Chan 1B]

Hi, you might want to double check your calculations. When I calculate standard enthalpy of formation for H2O(g), I got 4*-241.8 kJ/mol = -967.2 kJ/mol. This might be because you accidentally used the standard enthalpy of formation for H2O(l) instead.