Question

Calculate the standard entropy change for the reaction below at a temperature of 298 K:CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) The standard molar entropies are as follows: S(CH4) = 186.3 J K^-1 mol^-1S(O2) = 205.0 J K^-1 mol^-1S(CO2) = 213.7 J K^-1 mol^-1S(H2O) = 188.8 J K^-1 mol^-1

Answer 1

To calculate the standard entropy change for the reaction, we can use the following equation:S = S products  - S reactants For the reactants:1 mol of CH4: S CH4  = 186.3 J K^-1 mol^-12 mol of O2: S O2  = 2 * 205.0 J K^-1 mol^-1 = 410.0 J K^-1 mol^-1For the products:1 mol of CO2: S CO2  = 213.7 J K^-1 mol^-12 mol of H2O: S H2O  = 2 * 188.8 J K^-1 mol^-1 = 377.6 J K^-1 mol^-1Now, we can calculate the standard entropy change:S =  213.7 + 377.6  -  186.3 + 410.0 S = 591.3 - 596.3S = -5.0 J K^-1 mol^-1The standard entropy change for the reaction at 298 K is -5.0 J K^-1 mol^-1.