Consider the reaction at 298 K. C(graphite)+2H2(g)⟶CH4(g)Δ𝐻°=−74.6 kJ/mol Calculate the quantities for ΔSsys, ΔSsurr, ΔSuniv.

Answer 1

To calculate ΔSsys (change in entropy of the system), ΔSsurr (change in entropy of the surroundings), and ΔSuniv (change in entropy of the universe), we first need to know the reaction's ΔH (enthalpy change) and ΔG (Gibbs free energy change). We're given ΔH° = -74.6 kJ/mol at 298 K, and we'll assume the reaction is at standard conditions (1 atm and 298 K).

1. Calculate ΔSsys:

First, we need the standard entropy values (S°) of the reactants and products. We can find these values in standard entropy tables.

For this example, let's use the following values:

C(graphite): 5.7 J/mol·K H2(g): 130.7 J/mol·K CH4(g): 186.3 J/mol·K

Now, we can calculate the change in entropy for the system (ΔSsys) using the formula:

ΔSsys = ΣnS(products) - ΣnS(reactants)

ΔSsys = (1 × 186.3 J/mol·K) - (1 × 5.7 J/mol·K + 2 × 130.7 J/mol·K) = 186.3 - 267.1 = -80.8 J/mol·K

1. Calculate ΔSsurr:

To calculate the change in entropy of the surroundings (ΔSsurr), we'll use the formula:

ΔSsurr = -ΔH / T

Where ΔH is the enthalpy change and T is the temperature in Kelvin.

ΔSsurr = -(-74.6 × 10^3 J/mol) / 298 K = 74.6 × 10^3 / 298 = 250.3 J/mol·K

1. Calculate ΔSuniv:

Finally, we can calculate the change in entropy of the universe (ΔSuniv) by adding the changes in entropy of the system and surroundings:

ΔSuniv = ΔSsys + ΔSsurr = -80.8 J/mol·K + 250.3 J/mol·K = 169.5 J/mol·K

So, the quantities for ΔSsys, ΔSsurr, and ΔSuniv are:

ΔSsys = -80.8 J/mol·K ΔSsurr = 250.3 J/mol·K ΔSuniv = 169.5 J/mol·K