1. For each of the following reactions, calculate G rxn, H rxn,and S rxn. Rxn 1:
ID: 559691 • Letter: 1
Question
1. For each of the following reactions, calculate G rxn, H rxn,and S rxn.
Rxn 1: H2(g) + O2(g) = 2NO(g)
Rxn 2: 2NO(g) + O2(g) = 2NO2(g)
Rsn 3: 2O3(g) = 3O2(g)
Rxn 4: Na(s) =Na(g)
2. Which of the above reactions are exothermic? How do you know?
3 Which of the above reactions are spontaneous under standard conditions? How do you know?
4 Which of the above reactions would be made spontaneous by increasing the temperature of the reaction? Calculate the temperature needed to make it spontaneous.
5 Which of the above will always be spontaneous regardless of temperature?
6 Which will become non-spontaneous with increase of temperature?
For each of the following reactions, calculate AGerxn-arrrxn-and AS-xn- Hint: Use your text to help answer parts c-e. (Use extra paper) Reaction N2(g)+O2(g)2NO(g) 2NO(g)+O2 (g)2NO2(g) 203(g) 302(g) Na(s) Na(g) Rxn 1: Run 2: Rxn 3: Rxn 4: a. Which of the above reactions is/are exothermic? How do you know? b. Which of the above reactions is/are spontaneous under standard conditions? How do you know? Which of the above reactions could be made spontaneous by increasing the temperature of the reaction? Calculate the temperature necessary to make the reaction spontaneous c. d. Which of the above reactions will always be spontaneous, no matter what the temperature? Why? Which of the reactions above will become non-spontaneous by increasing the temperature of the reaction? Why? .Explanation / Answer
1. dHrxn, dSrxn and dGrxn
Rxn 1 : N2 (g) + O2 (g) --> 2NO (g)
dHrxn = dH(products) - dH(reactants)
= 2 x 90.2 = 180.4 kJ
dSrxn = dS(products) - dS(reactants)
= 2 x 210.7 - (191.5 + 205) = 24.9 J/K
taking T = 298 K
dGrxn = dHrxn - TdSrx
= 180.4 - 298 x 0.0249 = 172.98 kJ
Rxn 2 : 2NO (g) + O2 (g) ---> 2NO2 (g)
dHrxn = dH(products) - dH(reactants)
= 2 x 33.2 - (2 x 90.2) = -114 kJ
dSrxn = dS(products) - dS(reactants)
= 2 x 240 - (2 x 210.7 + 205) = -146.4 J/K
taking T = 298 K
dGrxn = dHrxn - TdSrx
= -114 - 298 x -0.1464 = -70.363 kJ
Rxn 3 : 2O3 (g) --> 3O2 (g)
dHrxn = dH(products) - dH(reactants)
= -2 x 142.7 = -285.4 kJ
dSrxn = dS(products) - dS(reactants)
= 3 x 205 - 2 x 238.9 = 137.2 J/K
taking T = 298 K
dGrxn = dHrxn - TdSrx
= -285.4 - 298 x 0.1372= -326.28 kJ
Rxn 4 : Na(s) --> Na(g)
dHrxn = dH(products) - dH(reactants)
= 108.8 kJ
dSrxn = dS(products) - dS(reactants)
= 153.6 - 51.2 = 102.4 J/K
taking T = 298 K
dGrxn = dHrxn - TdSrx
= 108.8 - 298 x 0.1024= 78.285 kJ
a. Rxn 2 and Rxn 3 are exothermic in nature as it has dHrxn -ve.
b. The reactions which are sponatenous under standad conditions have dGrxn as -ve value, that is Rxn 2 an Rxn 3.
c. Rxn 1 and Rxn 2 can be made spontaneous by increasing the temperature. Endothermic reaction rate increases by increasing temperature. This is according to the LeChatellier's principle.
Temperature for Rxn 1 to be spontaneous = 180.4/0.0249 = 7245.0 K
Temperature for Rxn 2 to be spontaneous = 108.8/0.1024 = 1062.5 K
d. Rxn 3 will always be sponatensou irrespective of temperature as both the dH and TdS terms are negative here.
e. Rxn 2 and Rxn 3 would becomes non-sponatneous by increasing the temperature as these are exothermic reactions, which according to LeChatellier's principle would go back towards reactants If temperature is increased to reestablish equilibrium. Excess heat given is taken by products to go to reactants until all of heat given is used up.
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