5 Qs Question 16 of 20 Question Assista Be sure to answer al

5 Qs Question 16 (of 20) Question Assista Be sure to answer all parts. View h G° for the reaction Questio H2(g) + 12(g) 2HI(g) Report a is 260 kJnual at 25°C, Calculate G, and predict the direction in which the reaction is spontaneous. The initial pressures are: PH, 3.40 atm P1.5 atmPIHI 1.75 atm k./mol The reaction is spontaneous in the forward direction The reaction is spontaneous in the reverse direction Cannot be determined. intel

Solution

dG0 = - 2.303RTlogK

2.6 * 10^3 = -2.303 * 8.314 * (25+298) *logK

Kc = 0.3798

Kp = Kc *(RT)^dn

here dn = 0 then Kp = Kc

              H2 (g) + I2 (g) <------> 2 HI (g)

initially     3.4       1.5             1.75

change         -x        -x              +2x

Equilibrium   (3.4-x)   (1.5-x)        (1.75+2x)

K = PHI^2/PH2*PI2

0.3798 = (1.75+2x)/((3.4-x)*(1.5-x))

x = 0.0487

at equilibrium PH2 = 3.4 - 0.0487 = 3.3513 atm

PI2 = 1.5 - 0.0487 = 1.4513 atm

PHI = 1.75+2(0.0487) = 1.8474 atm

Kp = PHI^2/PH2*PI2

Kp = (1.8474)^2/(3.3513*1.4513)

Kp = 0.7017

dG = dG0 + 2.303RTlogK

dG = 2.6 * 10^3 + (2.303 * 8.314 * 298 log(0.7017))

dG = 1722.164 J/mol

dG = 1.722 kJ/mol