Question

Consider the decomposition of barium carbonate: 2 Cr (s) + 3 O2 (g) 2CO3 (s) Using the following data, calculate AG at 25°C and determine if the reaction is spontaneous. Remember to convert to appropriate units. Give your answer in kj/mol*K. substance AS (/mol*KAH(kJ/mol*) Cr (s) 23.6 0 O2(g) 205.0 0 CrO3 (s) 81.2 -1139.7 AG Spontaneous at this temperature (25 °C)?

Answer 1

Change in Standard gibb's free energy change can be calculated by the formula;  ΔG⁰_rxn =  ΔH⁰_rxn - T (ΔS⁰rxn)

Where ΔH⁰ = Std. Change in enthalpy and ΔS⁰ = Std. Entropy change , T= Temperature

Given reaction is: 2Cr (s) + 3O₂ (g) -----> 2CrO₃ (g)

Substance	ΔS0 (J/mol-K)	ΔH0 (kJ/mol-K)
Cr(s)	23.5	0
O2(g)	205	0
CrO3(g)	81.2	-1139.7

From Hess's law: ΔS⁰ for the reaction 2Cr (s) + 3O₂ (g) -----> 2CrO₃ (g) can be given as

ΔS⁰_rxn = 2 ΔS⁰_(CrO3) - 2 ΔS⁰_Cr -3 ΔS⁰_O2

Putting the values of  ΔS⁰ we get

ΔS⁰_rxn = 2^(81.2 J/mol-K) - 2^(23.5 J/mol-K) - 3^(205 J/mol-K) = -499.15 J/mol-K

Also from Hess's law ΔH⁰_rxn is given by:

ΔH⁰_rxn = 2 ΔH⁰_CrO3 - 2 ΔH⁰_Cr - 3 ΔH⁰_O2

_---> ΔH⁰_rxn = 2^(-1139.7 kJ/mol-K) -2^(0) - 3^(0) =-2279.4 kJ/mol-K -2279400 J/mol-K

Now At T=25⁰C =25+273= 298K

ΔG⁰_rxn =  ΔH⁰_rxn - T (ΔS⁰_rxn)

---> ΔG⁰_rxn = (-2279400 J/mol-K) - (298)^(-499.15 J/mol-K) = 2130653.3 J/mol-K = -2130.653 kJ/mol-K

As Standard Gibb's free change of the given reaction comes to be negative. The reaction would be spontaneous at room temperature.

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