Calculate the standard change in Gibbs free energy of the following reactions at

Question

Calculate the standard change in Gibbs free energy of the following reactions at Standard Ambient Temperature and Pressure (SATP where T = 25 degree C and P = 1 atm) and label them as spontaneous or nonspontaneous. (a) 2 SO_2 + O_2 (g) right left arrow 2 SO_3 (g); delta H degree = -197.8 kJ, delta S degree = -188.0 J/K delta G degree = kJ, and the reaction is. (b) C_3H_8 (g) + 5 O_2 (g) right lift arrow 3 CO_2 (g) + 4 H_2 O (I); delta H degree = -2219.9 kJ, delta S degree = -374.9 J/K delta G degree = KJ, and the reaction is. C(diamond) right lift arrow C(graphite); delta H degree = - 1.9 KJ, delta S degree = + 3.3 J.K delta G degree = KJ, and the reaction is. SiO_2 (s) right lift arrow Si(s) + O_2(g); delta H degree = + 910.7 KJ, delta S degree = + 182.5 J/K delta G degree = KJ, and the reaction is.

Explanation / Answer

a)

delta H = -197.8KJ

delta S = -188.0J/k = -0.188 KJ/K

T = 25.0 oC =(25.0+ 273) K = 298.0 K

use:

delta G = delta H - T*delta S

delta G = -197.8 - 298.0 * -0.188

delta G = -141.8 KJ

since delta G is negative, it is spontaneous

b)

delta H = -2219.9KJ

delta S = -374.9J/k = -0.3749 KJ/K

T = 25.0 oC =(25.0+ 273) K = 298.0 K

use:

delta G = delta H - T*delta S

delta G = -2219.9 - 298.0 * -0.3749

delta G = -2108.2 KJ

since delta G is negative, it is spontaneous

c)

delta H = -1.9KJ

delta S = 3.3J/k = 0.0033 KJ/K

T = 25.0 oC =(25.0+ 273) K = 298.0 K

use:

delta G = delta H - T*delta S

delta G = -1.9 - 298.0 * 0.0033

delta G = -2.9 KJ

since delta G is negative, it is spontaneous

d)

delta H = 910.7KJ

delta S = 182.5J/k = 0.1825 KJ/K

T = 25.0 oC =(25.0+ 273) K = 298.0 K

use:

delta G = delta H - T*delta S

delta G = 910.7 - 298.0 * 0.1825

delta G = 856.3 KJ

since delta G is negative, it is non-spontaneous

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