Question

What is the entropy change (ΔS) for the reaction of burning 2 moles of methane gas (CH4) in excess oxygen gas (O2) at standard conditions? The balanced chemical equation for this reaction is:CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

Answer 1

To calculate the entropy change  S  for the reaction, we need the standard molar entropies  S  of the reactants and products. The standard molar entropies at 298 K are as follows:- CH4 g : 186.3 J/molK- O2 g : 205.2 J/molK- CO2 g : 213.8 J/molK- H2O g : 188.8 J/molKNow, we can calculate the entropy change using the formula:S = S products  - S reactants For the given reaction:S = [1 * 213.8 J/molK + 2 * 188.8 J/molK] - [1 * 186.3 J/molK + 2 * 205.2 J/molK]S = [213.8 + 377.6] - [186.3 + 410.4]S = 591.4 - 596.7S = -5.3 J/molKSince we have 2 moles of methane gas, the total entropy change for the reaction is:S_total = -5.3 J/molK * 2S_total = -10.6 J/molKSo, the entropy change  S  for the reaction of burning 2 moles of methane gas  CH4  in excess oxygen gas  O2  at standard conditions is -10.6 J/molK.