# Change in enthalpy and entropy for the formation of nickel tetracarbonyl

I was working on an enthalpy/entropy problem and I had a question on predicting the signs of some reactions. For the reaction

$$\ce{Ni (s) + 4 CO (g) -> Ni(CO)4 (g)}$$

my textbooks says that when we predict the sign change in enthalpy and entropy, enthalpy is > 0 and entropy < 0. I'm not sure why this is since bonds are forming in this reaction and energy is being released (negative sign in enthalpy) and when a solid and gas react to form gas, it increases disorder (positive sign in entropy).

Am I understanding something wrong? Why is enthalpy > 0 and entropy < 0?

• Wouldn't entropy be positive if the product is gaseous? – Joe Nov 19 '17 at 22:39
• Not sure about the enthalpy part, but the entropy is easily explained once you remember that you are transforming four moles of gas into one. – Jan Nov 20 '17 at 11:22
• What is the exact wording in your textbook? – Chet Miller Nov 20 '17 at 14:48