What is the change in entropy when 9.30 mL of liquid benzene (C6H6, d = 0.879 g/mL) is burned in the
presence of 37.2 L of oxygen gas, measured at 298 K and 1 atm pressure?
2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O(l); ∆S° = –437.7 J/K at 298 K
dSrxn = (n*dSo products) - (n*dSo reactants)
Look up the So values in your text/notes. They should be in a table at the end of your text.
To calculate the change in entropy (∆S) of the reaction, you can use the equation:
∆S = ΣnS°(products) - ΣmS°(reactants)
Where Σn represents the stoichiometric coefficients and S° represents the standard molar entropy.
In this case, we have the following stoichiometric coefficients for the reactants and products:
2C6H6(l), 15O2(g), 12CO2(g), 6H2O(l)
And the corresponding standard molar entropies (S°) for each substance:
S°(C6H6) = ?
S°(O2) = ?
S°(CO2) = ?
S°(H2O) = ?
To find the standard molar entropies for each substance, you can consult a reliable source such as a thermodynamic table or database. These values are usually given in units of J/(mol·K).
Once you have the values for S°, you can substitute them into the equation and calculate the change in entropy (∆S) for the reaction.
Please note that the density (d = 0.879 g/mL) of benzene is not directly used in the calculation of entropy. It is only provided as additional information about the substance.