Using the following thermochemical data, what is the change in enthalpy for the following reaction?
2H¿(g) + 2C(s) + O¿(g) d C¿HOH(l)?
C2H5OH(l) + 2O2(g) d 2CO2(g) + 2H2O(l), DH = -875 kJ
C(s) + O¿(g) d CO¿(g), DH = -394.51 kJ
H2(g) + ½O2(g) d H2O(l), DH = -285.8 kJ
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To determine the change in enthalpy for the given reaction, we can use the concept of Hess's Law. According to Hess's Law, the overall change in enthalpy for a reaction can be obtained by adding the enthalpy changes of the individual reactions that can be combined to give the desired reaction.
In this case, we need to rearrange the given reactions in order to obtain the desired reaction:
2H2(g) + 2C(s) + O2(g) → C2H5OH(l) + 2O2(g) + 2CO2(g) + 2H2O(l)
Now we can write the enthalpy change for the desired reaction by using the given enthalpy changes for the individual reactions. It is important to ensure that the coefficients balance when combining the equations.
First, we'll double the second equation to match the stoichiometry of O2 in the desired equation:
2C(s) + 2O2(g) → 2CO(g)
Then, we'll subtract the enthalpy change of the third equation (H2(g) + ½O2(g) → H2O(l)) from the enthalpy change of the first equation (2H2(g) + 2C(s) + O2(g) → C2H5OH(l) + 2O2(g) + 2CO2(g) + 2H2O(l)) because the third equation is in the reverse direction:
2H2(g) + 2C(s) + O2(g) → C2H5OH(l) + 2O2(g) + 2CO2(g) + 2H2O(l) - (H2(g) + ½O2(g) → H2O(l))
Now we can substitute the given values for the enthalpy changes:
DH = (-875 kJ) - (-285.8 kJ) = -589.2 kJ
Therefore, the change in enthalpy for the given reaction is -589.2 kJ.