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1. For the reaction 4 HCl(g) + O 2 (g) ---> 2 H 2 O(g) + 2 Cl 2 (g) Delta H° = -

ID: 1044766 • Letter: 1

Question

1. For the reaction

4HCl(g) + O2(g) ---> 2H2O(g) + 2Cl2(g)

Delta H° = -114.4 kJ and Delta S° = -128.9 J/K

The standard free energy change for the reaction of 2.44 moles of HCl(g) at 300 K, 1 atm would be ?? kJ.

This reaction is (reactant or product)  favored under standard conditions at 300 K. *choose either reactanct or product*

Assume that Delta H° and Delta S° are independent of temperature.

2. For the reaction

H2(g) + C2H4(g) ---> C2H6(g)

Delta G° = -105.9 kJ and Delta S° = -120.7 J/K at 258 K and 1 atm.

This reaction is (reactant or product) *chosse either reactant or product* favored under standard conditions at 258 K.

The standard enthalpy change for the reaction of 1.83 moles of H2(g) at this temperature would be ?? kJ.


3. For the reaction

NH4NO3(aq) ---> N2O(g) + 2 H2O(l)

Delta G° = -178.9 kJ and Delta H° = -149.6 kJ at 293 K and 1 atm.

This reaction is (reactant, product) *choose either reactant ot product* favored under standard conditions at 293 K.

The entropy change for the reaction of 1.84 moles of NH4NO3(aq) at this temperature would be ?? J/K.

Explanation / Answer

1.   4HCl(g) + O2(g) ---> 2H2O(g) + 2Cl2(g)

Delta H° = -114.4 kJ and Delta S° = -128.9 J/K

DH = -114.4*2.44/4 = -69.8 kj

DS0 = -128.9*2.44/4 = -78.63 j

T = 300 k

DG = DH-TDS

    = -69.8 -(300*-78.63*10^-3)

    = -46.21 Kj

As DG = - ve, the reaction is product favoured.


2.   H2(g) + C2H4(g) ---> C2H6(g)

Delta G° = -105.9 kJ and Delta S° = -120.7 J/K at 258 K and 1 atm.

As DG = - ve, the reaction is product favoured.

DG = DH-TDS

-105.9 = x-(258*-120.7*10^-3)

x = DHrxn = -137.04 kj/mol

per 1.83 mol DHrxn = - 137.04*1.83/1 = -250.8 kj

answer: -250.8 kj

3.

NH4NO3(aq) ---> N2O(g) + 2 H2O(l)

Delta G° = -178.9 kJ and Delta H° = -149.6 kJ at 293 K and 1 atm.

As DG = - ve, the reaction is product favoured.

DG = DH-TDS

-178.9 = (-149.6) - 293*x*10^-3

x = DS= 100 j/k.mol

per 1.84 mol NH4NO3(aq) ,

DS = 100*1.84/1 = 184 j/k

answer: 184 j/k