Question

Calculate the change in entropy for the combustion of methane gas (CH4) at constant pressure and temperature of 298 K. The products of the reaction are carbon dioxide gas (CO2) and water vapor (H2O). The molar entropy of CH4, CO2 and H2O at 298 K are 186.3 J/K/mol, 213.7 J/K/mol, and 188.8 J/K/mol, respectively.

Answer 1

To calculate the change in entropy for the combustion of methane gas, we need to consider the balanced chemical equation for the reaction:CH g  + 2O g   CO g  + 2HO g Next, we will use the formula for the change in entropy  S  of a reaction:S = nS products  - nS reactants where n is the stoichiometric coefficient and S is the molar entropy of each species.For the products:1 mol of CO: 1  213.7 J/K/mol = 213.7 J/K2 mol of HO: 2  188.8 J/K/mol = 377.6 J/KFor the reactants:1 mol of CH: 1  186.3 J/K/mol = 186.3 J/K2 mol of O: Since the problem does not provide the molar entropy of O, we will assume that the change in entropy of O is negligible. This is a reasonable assumption because O is a stable molecule and its entropy change should not significantly affect the overall entropy change of the reaction.Now, we can calculate the change in entropy  S :S =  213.7 + 377.6  -  186.3  = 591.3 - 186.3 = 405 J/KThe change in entropy for the combustion of methane gas at constant pressure and temperature of 298 K is 405 J/K.