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Among other things, an ideal fuel for the control thrusters of a space vehicle should decompose in a spontaneous exothermic reaction when exposed to the appropriate catalyst. Evaluate the following substances under standard state conditions as suitable candidates for fuels.

(a) Ammonia: 2NH 3 ( g ) N 2 ( g ) + 3H 2 ( g )

(b) Diborane: B 2 H 6 ( g ) 2B ( g ) + 3H 2 ( g )

(c) Hydrazine: N 2 H 4 ( g ) N 2 ( g ) + 2H 2 ( g )

(d) Hydrogen peroxide: H 2 O 2 ( l ) H 2 O ( g ) + 1 2 O 2 ( g )

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Calculate Δ G ° for each of the following reactions from the equilibrium constant at the temperature given.

(a) N 2 ( g ) + O 2 ( g ) 2NO ( g ) T = 2000 °C K p = 4.1 × 10 −4

(b) H 2 ( g ) + I 2 ( g ) 2HI ( g ) T = 400 °C K p = 50.0

(c) CO 2 ( g ) + H 2 ( g ) CO ( g ) + H 2 O ( g ) T = 980 °C K p = 1.67

(d) CaCO 3 ( s ) CaO ( s ) + CO 2 ( g ) T = 900 °C K p = 1.04

(e) HF ( a q ) + H 2 O ( l ) H 3 O + ( a q ) + F ( a q ) T = 25 °C K p = 7.2 × 10 −4

(f) AgBr ( s ) Ag + ( a q ) + Br ( a q ) T = 25 °C K p = 3.3 × 10 −13

(a) 1.5 × 10 2 kJ; (b) −21.9 kJ; (c) −5.34 kJ; (d) −0.383 kJ; (e) 18 kJ; (f) 71 kJ

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Calculate Δ G ° for each of the following reactions from the equilibrium constant at the temperature given.

(a) Cl 2 ( g ) + Br 2 ( g ) 2BrCl ( g ) T = 25 °C K p = 4.7 × 10 −2

(b) 2SO 2 ( g ) + O 2 ( g ) 2SO 3 ( g ) T = 500 °C K p = 48.2

(c) H 2 O ( l ) H 2 O ( g ) T = 60 °C K p = 0.196 atm

(d) CoO ( s ) + CO ( g ) Co ( s ) + CO 2 ( g ) T = 550 °C K p = 4.90 × 10 2

(e) CH 3 NH 2 ( a q ) + H 2 O ( l ) CH 3 NH 3 + ( a q ) + OH ( a q ) T = 25 °C K p = 4.4 × 10 −4

(f) PbI 2 ( s ) Pb 2+ ( a q ) + 2I ( a q ) T = 25 °C K p = 8.7 × 10 −9

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Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of Δ G ° given.

(a) O 2 ( g ) + 2F 2 ( g ) 2OF 2 ( g ) Δ G ° = −9.2 kJ

(b) I 2 ( s ) + Br 2 ( l ) 2IBr ( g ) Δ G ° = 7.3 kJ

(c) 2LiOH ( s ) + CO 2 ( g ) Li 2 CO 3 ( s ) + H 2 O ( g ) Δ G ° = −79 kJ

(d) N 2 O 3 ( g ) NO ( g ) + NO 2 ( g ) Δ G ° = −1.6 kJ

(e) SnCl 4 ( l ) SnCl 4 ( l ) Δ G ° = 8.0 kJ

(a) K = 41; (b) K = 0.053; (c) K = 6.9 × 10 13 ; (d) K = 1.9; (e) K = 0.04

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Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of Δ G ° given.

(a) I 2 ( s ) + Cl 2 ( g ) 2ICl ( g ) Δ G ° = −10.88 kJ

(b) H 2 ( g ) + I 2 ( s ) 2HI ( g ) Δ G ° = 3.4 kJ

(c) CS 2 ( g ) + 3Cl 2 ( g ) CCl 4 ( g ) + S 2 Cl 2 ( g ) Δ G ° = −39 kJ

(d) 2SO 2 ( g ) + O 2 ( g ) 2SO 3 ( g ) Δ G ° = −141.82 kJ

(e) CS 2 ( g ) CS 2 ( l ) Δ G ° = −1.88 kJ

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Calculate the equilibrium constant at the temperature given.

(a) O 2 ( g ) + 2F 2 ( g ) 2F 2 O ( g ) ( T = 100 °C )

(b) I 2 ( s ) + Br 2 ( l ) 2IBr ( g ) ( T = 0.0 °C )

(c) 2LiOH ( s ) + CO 2 ( g ) Li 2 CO 3 ( s ) + H 2 O ( g ) ( T = 575 °C )

(d) N 2 O 3 ( g ) NO ( g ) + NO 2 ( g ) ( T = −10.0 °C )

(e) SnCl 4 ( l ) SnCl 4 ( g ) ( T = 200 °C )

In each of the following, the value of Δ G is not given at the temperature of the reaction. Therefore, we must calculate Δ G from the values Δ H ° and Δ S and then calculate Δ G from the relation Δ G = Δ H ° − T Δ S °.
(a) K = 1.29;
(b) K = 2.51 × 10 −3 ;
(c) K = 4.83 × 10 3 ;
(d) K = 0.219;
(e) K = 16.1

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Calculate the equilibrium constant at the temperature given.

(a) I 2 ( s ) + Cl 2 ( g ) 2ICl ( g ) ( T = 100 °C )

(b) H 2 ( g ) + I 2 ( s ) 2HI ( g ) ( T = 0.0 °C )

(c) CS 2 ( g ) + 3Cl 2 ( g ) CCl 4 ( g ) + S 2 Cl 2 ( g ) ( T = 125 °C )

(d) 2SO 2 ( g ) + O 2 ( g ) 2SO 3 ( g ) ( T = 675 °C )

(e) CS 2 ( g ) CS 2 ( l ) ( T = 90 °C )

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Consider the following reaction at 298 K:
N 2 O 4 ( g ) 2NO 2 ( g ) K P = 0.142

What is the standard free energy change at this temperature? Describe what happens to the initial system, where the reactants and products are in standard states, as it approaches equilibrium.

The standard free energy change is Δ G 298 ° = R T ln K = 4.84 kJ/mol . When reactants and products are in their standard states (1 bar or 1 atm), Q = 1. As the reaction proceeds toward equilibrium, the reaction shifts left (the amount of products drops while the amount of reactants increases): Q <1, and Δ G 298 becomes less positive as it approaches zero. At equilibrium, Q = K , and Δ G = 0.

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Practice Key Terms 3

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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