"Translate the empirical molar enthalpy given below into a balanced chemical equation, including the enthalpy change (delta H):
The standard molar enthalpy of formation for carbon disulfide is 89.0 kJ/mol"
I got delta H = 178 kJ (is that right?). However, I'm having trouble knowing whether it's + or -.
C+2S>>CS2 Hf=+89.0kj/mole
To determine the sign of the enthalpy change (ΔH), you need to know whether the reaction is exothermic or endothermic. An exothermic reaction releases heat, so the ΔH value would be negative. On the other hand, an endothermic reaction absorbs heat, so the ΔH value would be positive.
In this case, you are given the standard molar enthalpy of formation for carbon disulfide, which indicates the formation of one mole of carbon disulfide under standard conditions. The standard enthalpy of formation is always for the formation of products, not the reactants.
The standard molar enthalpy of formation for carbon disulfide is 89.0 kJ/mol. To write the balanced chemical equation, we need to know the reactants and the products. Assuming that sulfur (S8) and carbon (C) are the reactants and carbon disulfide (CS2) is the product, the balanced chemical equation would be:
S8 + 4C → 4CS2
Note that the coefficient of carbon disulfide is 4 because the standard molar enthalpy of formation refers to moles of the product formed.
Now, to determine the sign of the enthalpy change (ΔH), we need to look at the formation of carbon disulfide. We can interpret the reaction as follows:
4 moles of CS2 are formed using sulfur and carbon as reactants. The reaction is the formation of CS2, which means heat is released (exothermic). Therefore, the enthalpy change (ΔH) for this reaction is negative:
ΔH = -89.0 kJ/mol