At the boiling point the liquid and vapor states are in equilibrium. What is the

Question

At the boiling point the liquid and vapor states are in equilibrium. What is the boiling point of benzene if Delta H_vap = 30.7 kJ/mol and Delta S_vap = 87.0 J/mol-K.? Calculate Delta G degree at 298 K for the equation below using the thermodynamic data given: N_2(g) + 3 H_2(g) rightarrow 2 NH_3(g) Delta H degree (NH_3) = -46 kJ: Delta S degree values (J/mol-K) are NH_3 = 192.5, N_2 = 191.5, H_2 130.6. Determine Delta G for the following reaction at 500 K and 0.5 atm for: CH_4(g) + 2Cl_2(g) rightarrow CCl_4(g) + 2H_2(g) (At 298K: Delta H degree = - 28.09 kJ/mol, Delta S degree = -61.20 kJ/mol-K)

Explanation / Answer

Q7

Tboiling --> liq-vap...

dHvap = 30.7 kJ/mol = 30700 J/mol

dSvap = 87 J/mol K

so..

dG = 0 , in equilibrium

so

dG = dH - dTS = 0

so

dH = dS*T

T = dH/dS

T = 30700 / 87

T = 352.873 K

T = 79.873 °C

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