Question

Calculate the entropy change for the combustion of methane (CH4) at 298 K and 1 atm pressure, given the standard molar entropies of CH4, H2O (g), and O2 (g) are 186.3 J/K/mol, 188.7 J/K/mol, and 205.0 J/K/mol, respectively. The balanced chemical equation for the reaction is:CH4 (g) + 2 O2 (g) -> CO2 (g) + 2 H2O (g)

Answer 1

To calculate the entropy change  S  for the combustion of methane, we need to use the standard molar entropies of the reactants and products involved in the reaction. The balanced chemical equation for the reaction is:CH4  g  + 2 O2  g  -> CO2  g  + 2 H2O  g The entropy change for the reaction can be calculated using the following formula:S = nS products  - nS reactants where n is the stoichiometric coefficient and S is the standard molar entropy.For the reactants:- 1 mol of CH4  g  has a standard molar entropy of 186.3 J/K/mol- 2 mol of O2  g  have a standard molar entropy of 205.0 J/K/mol eachFor the products:- 1 mol of CO2  g  has a standard molar entropy of 213.7 J/K/mol  not provided in the problem, but can be found in standard chemistry references - 2 mol of H2O  g  have a standard molar entropy of 188.7 J/K/mol eachNow, we can calculate the entropy change:S = [ 1  213.7  +  2  188.7 ] - [ 1  186.3  +  2  205.0 ]S =  213.7 + 377.4  -  186.3 + 410.0 S = 591.1 - 596.3S = -5.2 J/KThe entropy change for the combustion of methane at 298 K and 1 atm pressure is -5.2 J/K.