Chemistry Forum

1]Consider the following balanced reaction:

\(2A_2 + 3B \rightarrow C_2 + 2D \)


What is the enthalpy of reaction if the enthalpy of formation of the compounds is given as
\(A_2 = 7.78 kJ/mol; B = -18.43 kJ/mol; C_2 = 0 kJ/mol; D = 9.96 kJ/mol.\)

Record your answer in scientific notation using 3 significant figures.

My answer to this question is 59.65 kJ/mol computed as follows: [9.96*2-2*(7.78)-3*(-18.43)=59.65]

2]Calculate ΔH° for the process Co₃O₄(s) → 3 Co(s) + 2 O₂(g)

from the following information

Co(s) + ½ O₂(g) → CoO(s) ΔH° = -120.6 kJ/mol
3 CoO(s) + ½ O₂(g) → Co₃O₄(s) ΔH° = -632.4 kJ/mol


I am working on these two questions. Any chemistry help will be accepted.

The first one is -59.65 kJ/ mol educt- product.

The second one 3*(-120,6 kJ/mol) + (-632,4 kJ/mol) = -994,2 kJ/ mol

Both the reactions are endothermic , but your negative signs implies both reactions are exothermic. How is that?

Who said that. In first it was told your result is wrong.
So take the equation half, the Addition of educt is one enthalpy value. The product also have value, what has subtracted vom educt value, it gives negative value.

The cobaltoxide reaction.

First CoO is formed with negativ value, from there Co₃O₄ is formed also with negativ value. The summery is still negativ. But it is the decomposition asked, that it is then the opposit positive.

Can I ask an off-topic question? I want to buy an infrared thermometer but don't know a good brand, can you advise me? Thanks.