Question

Calculate the entropy change (ΔS) for the combustion of methane gas (CH4) at constant pressure and standard conditions, given the following data:   C(s) + O2(g) → CO2(g)      ΔS° = -394 J/K·mol   2H2(g) + O2(g) → 2H2O(l)   ΔS° = -237 J/K·mol   CH4(g) + 2O2(g) → CO2(g) + 2H2O(l)   ΔH° = -802 kJ/mol

Answer 1

To calculate the entropy change  S  for the combustion of methane gas  CH4  at constant pressure and standard conditions, we can use the Hess's Law. According to Hess's Law, the total entropy change for a reaction is the sum of the entropy changes for the individual steps that make up the overall reaction.First, we need to rewrite the given reactions in a way that they add up to the desired reaction:1. C s  + O2 g   CO2 g       S = -394 J/Kmol2. 2H2 g  + O2 g   2H2O l    S = -237 J/KmolThe desired reaction is:CH4 g  + 2O2 g   CO2 g  + 2H2O l We can break down the desired reaction into two steps:Step 1: CH4 g   C s  + 2H2 g Step 2: C s  + 2H2 g  + 2O2 g   CO2 g  + 2H2O l Now, we need to find the entropy change for Step 1. Since we don't have the entropy change for this reaction, we can use the fact that S = H/T, where H is the enthalpy change and T is the temperature. We know that H for the desired reaction is -802 kJ/mol, and the standard temperature is 298 K. Therefore, S for Step 1 can be calculated as:S =  -802 kJ/mol  / 298 K = -2.694 J/KmolNow, we can calculate the total entropy change for the desired reaction by adding the entropy changes for Step 1 and Step 2:S_total = S_step1 + S_step2S_total = -2.694 J/Kmol +  -394 J/Kmol  +  -237 J/Kmol S_total = -633.694 J/KmolSo, the entropy change  S  for the combustion of methane gas  CH4  at constant pressure and standard conditions is -633.694 J/Kmol.