Question

Find ∆°H for the reaction 2H2(g) + 2C(s) + O2(g) C2H5OH(l) using the following thermochemical equations: (4 marks) C2H5OH(l) + 2O2(g) 2CO2(g) + 2H2O(l) ∆H° = - 875 kJC(s) + O2(g) CO2(g) ∆H°= -394.51 kJH2(g) + ½ O2(g) H2O(l) ∆H° = - 285.8 kJ

Answer 1

Step-by-step explanation:

Here, we want to get the ∆°H for the reaction

Mathematically, to get this, we have to subtract the change in enthalpy of the reactants from that of the product

However, we shall not be considering the enthalpy values for single element molecules like hydrogen molecule and oxygen molecule

From the later reactions and values given, we can get the values for carbon (iv) oxide and Ethanol

For Carbon (iv) oxide, we have it as:

`\begin{gathered} \text{ enthalpy of product - enthalpy of reactant =-394.5 KJ} \\ enthalpy\text{ of reactant here is zero} \\ \text{ Thus enthalpy of product CO}_2\text{ = -394.51 KJ} \end{gathered}`

For Ethanol, we need to get the value for water

We have that as follows:

`-285.8\text{ KJ}`

For ethanol, that would be: Let us label it e