Using the following thermochemical data, what is the change in enthalpy for the following reaction: 3h2(g) + 2c(s) + ½o2(g) c2h5oh(l) , δh = –1367 kj/mol , ½ ,

Given: 

3H2(g) + 2C(s) + ½O2(g)     →   C2H5OH(l) , ΔH = –1367 kJ/mol 

Change is enthalpy is mathematically expressed as enthalpy of product minus enthalpy of reactant. If value of enthalpy change is negative, it suggest reaction is exothermic. If enthalpy change is positive, it suggest reaction is endothermic.

In present case, enthalpy change is -1367 kJ/mol.

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