Diethyl ether, \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l),\) a flammable compound that has long been used as a surgical anesthetic, has the structure \(\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2}-\mathrm{CH}_{3}\) The complete combustion of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) yields \(\Delta H^{\circ}=-2723.7 \mathrm{~kJ}\). (a) Write a balanced equation for the combustion of \(1 \mathrm{~mol}\) of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) (b) Write a balanced equation for the formation of \(\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}(l)\) from its elements. (c) By using the information in this problem and data in Table \(5.3,\) calculate \(\Delta H_{f}^{\circ}\) for diethyl ether.

To write the balanced equation for the combustion of 1 mol of diethyl ether (C4H10O), we need to know that: - Complete combustion means the reactants are Diethyl ether and O2; the products are CO2 and H2O. - The coefficients should be adjusted so that the atoms are conserved. The unbalanced equation looks like this: C4H10O(l) + O2(g) → CO2(g) + H2O(l) Now, balance the equation by adjusting the coefficients: C4H10O(l) + \(6\frac{1}{2}\) O2(g) → 4 CO2(g) + 5 H2O(l)

To write the balanced equation for the formation of diethyl ether from its elements, we need to know that: - Formation means the reactants are the elements in their standard states; the product is diethyl ether. - The coefficients should be adjusted so that the atoms are conserved. The unbalanced equation looks like this: C + H2 + O2 → C4H10O Now, balance the equation by adjusting the coefficients: 4 C + 5 H2 + \( \frac{1}{2} \) O2 → C4H10O

To calculate the standard enthalpy of formation for diethyl ether, use the given information and the standard enthalpy of formation for the involved species from Table 5.3. We use the following equation: ∆Hf°(C4H10O) = ∆H°(products) - ∆H°(reactants) First, gather the standard enthalpy of formation values from table 5.3: ∆Hf°(C) = 0 kJ/mol (as carbon is in its standard state) ∆Hf°(H2) = 0 kJ/mol (as hydrogen is in its standard state) ∆Hf°(O2) = 0 kJ/mol (as oxygen is in its standard state) ∆Hf°(CO2) = -393.5 kJ/mol ∆Hf°(H2O) = -285.8 kJ/mol Now, we calculate the enthalpy change for the combustion of C4H10O from the given information: ∆H°(combustion) = -2723.7 kJ According to the balanced combustion equation (Step 1): ∆H°(combustion) = 4 ∆Hf°(CO2) + 5 ∆Hf°(H2O) - ∆Hf°(C4H10O) -2723.7 kJ = 4(-393.5 kJ/mol) + 5(-285.8 kJ/mol) - ∆Hf°(C4H10O) Now, solve for ∆Hf°(C4H10O): ∆Hf°(C4H10O) = 4(-393.5 kJ/mol) + 5(-285.8 kJ/mol) + 2723.7 kJ ∆Hf°(C4H10O) = - 261.6 kJ/mol The standard enthalpy of formation for diethyl ether is -261.6 kJ/mol.