1.16 At 298.15 K, all of the following substances have a standard free energy of

Question

1.16 At 298.15 K, all of the following substances have a standard free energy of formation of zero EXCEPT ____. a) Br2(l) b) I2(s) c) S8(s) d) Cl2(g) e) Hg(g)

1.18
The standard free energy change for a chemical reaction is +21.9 kJ/mole. What is the equilibrium constant for the reaction at 87 1.16 At 298.15 K, all of the following substances have a standard free energy of formation of zero EXCEPT ____. a) Br2(l) b) I2(s) c) S8(s) d) Cl2(g) e) Hg(g)

1.18
The standard free energy change for a chemical reaction is +21.9 kJ/mole. What is the equilibrium constant for the reaction at 87 The standard free energy change for a chemical reaction is +21.9 kJ/mole. What is the equilibrium constant for the reaction at 87

Explanation / Answer

1.16] e is the answer

REASON : standard stste of Hg is liquid but its given Hg(g) and substance has zero free energy of formation only for standard state

1.18] free energy change delta G = -RTlnKc

            where Kc is the equilibrium constant

so susbtitiuting values given   T = 87 C =360 K

21900 =- 8.314*360 ln Kc

Kc =6.642*10^(-4)     ANS

1.19] Kc = 6.642*10^(-4)    

1.20] given     reaction CCl4(g) ------> CCl4(l)

CCl4(g)    CCl4(l)          
/H (kJ/mol-rxn)        -96    -128.4           
S (J/Kmolrxn)         309.9    214.4          
/ G (kj/molrxn)   -53.6 -57.6           

use delta G = delta H - T* delta S   where T is temp of condensation

delta H = H of product - H of reactant = -128.4 -[-96] =-32.4 kJ = -32400 J/mol

delta G = G of product - G of reactant= -57.6   -[-53.6] = -4 kJ = -4000 J/mol

delta S= S of product - S of reactant = 214.4- 309.9 =-95.5 J/ mol

so delta G = delta H - T* delta S

-4000 = -32400 - T*(-95.5)

gives T = 297.38 K

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