For complete combustion of methanol \( CH_3OH(I)+\frac 32O_2(g)→CO_2(g)+2H_2O(I)\) the amount of heat produced as measured by bomb calorimeter is \(726\) kJ mol–1 at \(27°C\). The enthalpy of combustion for the reaction is \(x \ kJ mol^{–1}\), where \(x\) is ______ . (Nearest integer) (Given: \(R= 8.3 \ JK^{–1}mol^{–1}\))
The enthalpy change \( \Delta H \) is related to the internal energy change \( \Delta U \) by the equation \(\Delta H = \Delta U + \Delta nRT\), where \( \Delta n \) is the change in moles of gas, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
For the given reaction: \(\begin{align*} CH_3OH(l) + \frac{3}{2}O_2(g) & \rightarrow CO_2(g) + 2H_2O(l) \end{align*}\) \(\Delta n = n_{\text{products}} - n_{\text{reactants}} = 1 - \frac{3}{2} = -\frac{1}{2}\). Temperature conversion: \(27°C = 300K\). Substitute in the equation: \(\Delta H = -726 \, \text{kJ/mol} + \left(-\frac{1}{2}\right) \times 8.3 \, \text{J/K/mol} \times 300 \, \text{K}\). Calculate the correction term: \(-\frac{1}{2} \times 8.3 \times 300 = -1245 \, \text{J/mol} = -1.245 \, \text{kJ/mol}\). Thus, \(\Delta H = -726 \, \text{kJ/mol} - 1.245 \, \text{kJ/mol} = -727.245 \, \text{kJ/mol}\). Rounded to the nearest integer, \( x = 727 \, \text{kJ/mol}\). This fits the expected range of 727,727.
