Question

Calculate the entropy change for the reaction of 2 moles of methane gas reacting with 4 moles of oxygen gas to form carbon dioxide gas and water vapor at a constant temperature of 298 K. The reaction is balanced and the standard entropy values for methane, oxygen, carbon dioxide, and water vapor are given as -0.242, 205.16, 213.74, and 188.72 J/K/mol, respectively.

Answer 1

To calculate the entropy change for the reaction, we can use the following equation:S_reaction = n_products * S_products - n_reactants * S_reactantswhere S_reaction is the entropy change for the reaction, n is the number of moles of each substance, and S is the standard entropy value for each substance.The balanced reaction is:2 CH4  g  + 4 O2  g   2 CO2  g  + 4 H2O  g Now, we can plug in the values for the moles and standard entropy values for each substance:S_reaction = [ 2 * 213.74  +  4 * 188.72 ] - [ 2 *  -0.242   +  4 * 205.16 ]S_reaction = [ 427.48  +  754.88 ] - [ 0.484  +  820.64 ]S_reaction = 1182.36 - 821.124S_reaction = 361.236 J/KThe entropy change for the reaction of 2 moles of methane gas reacting with 4 moles of oxygen gas to form carbon dioxide gas and water vapor at a constant temperature of 298 K is 361.236 J/K.