Please help me find delta G! T College Chemistry -II Summer 2015 CH 17 HW Exerci

Question

Please help me find delta G! T College Chemistry -II Summer 2015 CH 17 HW Exercise 17.72 with feedback Consider the following reaction: 12(g) + Cl2 (g) 21C1(g) K,-81.9 at 25 °C Calculate Gnn for the reaction at 25° C under each of the following conditions. | | Correct You may want to reference (LD pages 842-845 section 17.8 while completing this problem Under equilbrium conditions, the value of RT InQ is aheas equal in nagnitude but oppos sign to the value of AG spontaneous in ether direcion, as expected for a neacion at equilbrium Part C =263atm : | = 0.324 atm P0 221 atm Express your answer using one signficant figure R S Incorrect: Try Again; 5 attempts

Explanation / Answer

Answer –

In this problem we are given reaction-

I2(g) + Cl2(g) <-----> 2 ICl(g) Kp = 81.9

P ICl = 2.63 atm

P I2 = 0.324 atm

P Cl2 = 0.221 atm

First we need to calculate the standard free energy from the given Kp

Go = -RTlnK

       = - 8.314 J.mol-1.K-1 * 298 K * ln 81.9

       = -10915 J

Now e know the formula for calculating the free energy

G = Go + RTln Q

     = (-10915 J) + ( 8.314 J.mol-1.K-1 * 298 K * ln [(P ICl)2 / (PCl2) *(P I2)]

    = (-10915 J) + ( 8.314 J.mol-1.K-1 * 298 K * ln [(0.221)2 / (2.63*0.324)

    = -10915 J – 7083.77 J

    = -17998.9 J

    = - 17.99 kJ

We need answer in one sig fig, so -17.99 is nothing but -20 kJ in one sig fig.

So answer is -20 kJ

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.