The molar enthalpies of combustion of CH3COCOOH(l)CH3COOH(l) and CO(g) are respectively -1275kJ/mol, -875kJ/mol, and -283kJ/mol. What is the enthalpy change for the reaction below?
CH3COCOOH= CH3COOH +CO
a)1867kJ
b)-1867kJ
c)117kJ
d)-117kJ
e)-2433kJ
I think it's d)
Did you check your work as I suggested earlier? I still don't get -117 kJ. Post your work and let me see if I can find the error.
I thought what your suppose to do is
-1275 - ((-875) + (-283))= -117kj/mols
Don't u just subtract? To find the difference?
delta Hrxn = (sum DH products)-(sum DH reactants)
DHrxn = (-873-283)-(-1275)
-1158 + 1275 = +117 kJ.
Can you explain why you have to subtract the other way?
To find the enthalpy change for the reaction, you need to use the Hess's Law, which states that the overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for each step of the reaction.
In this case, you have the following steps:
1. CH3COCOOH(l) → CH3COOH(l) + CO(g) (Given reaction)
2. CH3COOH(l) + CO(g) → CH3COCOOH(l) (Reverse of the given reaction)
To find the enthalpy change for the given reaction, you can add the enthalpy changes of the individual steps. The enthalpy change for the reverse reaction will have the opposite sign.
Given:
Enthalpy change for step 1: -1275 kJ/mol
Enthalpy change for step 2: -(-283 kJ/mol) = 283 kJ/mol (reverse of -283 kJ/mol)
Now, add the enthalpy changes of both steps:
Enthalpy change for the given reaction = Enthalpy change for step 1 + Enthalpy change for step 2
= -1275 kJ/mol + 283 kJ/mol
= -992 kJ/mol
Notice that the value is negative, indicating an exothermic reaction. Therefore, the correct answer is option d) -117 kJ, which is the negative of -992 kJ (answer).
So, you are correct! The enthalpy change for the given reaction is -117 kJ.