calculate the change in entropy when liquid mercury is vaporized.
Hg(l) = 75.9 J/mol K Hg(g) = 175.0 J/mol K
Hg(l) + heat ==> Hg(g)
dSrxn = dS(products) - dS(reactants)
dSrxn = (1 mol Hg(g) x 175.0 J/mol*K) - (1 mol Hg(l) x 75.9 J/mol*K) = ?
To calculate the change in entropy (ΔS) when liquid mercury is vaporized, we can use the formula:
ΔS = S(g) - S(l)
Where S(g) is the molar entropy of the gaseous state and S(l) is the molar entropy of the liquid state.
Given the molar entropies of liquid mercury (Hg(l)) and gaseous mercury (Hg(g)), we have:
S(l) = 75.9 J/mol K
S(g) = 175.0 J/mol K
Substituting these values into the formula, we can calculate the change in entropy:
ΔS = 175.0 J/mol K - 75.9 J/mol K
= 99.1 J/mol K
Therefore, the change in entropy when liquid mercury is vaporized is 99.1 J/mol K.