Consider the sublimation of iodine at 25.0 °C. I2(s) I2(g) a. Find G°rxn at 25.0
ID: 496745 • Letter: C
Question
Consider the sublimation of iodine at 25.0 °C. I2(s) I2(g) a. Find G°rxn at 25.0 °C. b. Find Grxn at 25.0 °C under the following nonstandard conditions: i. PI2 = 1.00 mmHg ii. PI2 = 0.100 mmHg c. Explain why iodine spontaneously sublimes in open air at 25 °C.2) Balance the following redox reaction in basic aqueous solution. Cl2(g) Cl– (aq) + ClO– (aq) Consider the sublimation of iodine at 25.0 °C. I2(s) I2(g) a. Find G°rxn at 25.0 °C. b. Find Grxn at 25.0 °C under the following nonstandard conditions: i. PI2 = 1.00 mmHg ii. PI2 = 0.100 mmHg c. Explain why iodine spontaneously sublimes in open air at 25 °C.
2) Balance the following redox reaction in basic aqueous solution. Cl2(g) Cl– (aq) + ClO– (aq) Consider the sublimation of iodine at 25.0 °C. I2(s) I2(g) a. Find G°rxn at 25.0 °C. b. Find Grxn at 25.0 °C under the following nonstandard conditions: i. PI2 = 1.00 mmHg ii. PI2 = 0.100 mmHg c. Explain why iodine spontaneously sublimes in open air at 25 °C.
2) Balance the following redox reaction in basic aqueous solution. Cl2(g) Cl– (aq) + ClO– (aq)
Explanation / Answer
now delta Go = 19.3 kj = 19.3 X 1000 = 19300 j
delta G = delta Go + RTlnQ
when PI2 = 1 mm Hg
Q = PI2 = 1 mm/760 atm ( as 1 atm = 760 mm and partial pressure should be in atm in reaction quotient
expression)
R = 8.314 j/K/mole
T = 298 K
so delta G = 19300 + 8.314 X 298 X ln (1/760)
delta G = 19300 - 16434.524 = 2865.476 j or 2865.476/1000 kj = 2.86 kj = 2.9 kj
which is your answer
similarly when PI2 = 0.1 mm
dlta G = 19300 + 8.314 X 298 X ln ( 0.1/760)
dlta G = 19300 - 22139.344 = -2839.344 j or -2839.344/1000 = -2.83 kj
which is almost equal to your answer
The phase diagrams of various substances vary. Iodine sublimes because even though it has significant van-der-waals (intermolecular) forces that could hold it as a solid, it has very high vapour pressure and any liquid that is formed becomes gaseous right away at 1 atm. Thus it sublimes at open air.
2) Cl2 (g) Oxidation number of 0
Cl- (aq) Oxidation number of -1
ClO- (aq) Cl is +1 and O is -2.
We do not need to multiply anything by anything. The problem is where did the Oxygen as wekk as the charge come from? Our choices in a BASIC solution are H2O or OH-. Since OH- is the only one charged, add two of them to the left side to balance the charges.
Cl2 (g) + 2 OH- (aq) ---> Cl- (aq) + ClO- (aq)
OK, Cl's still balance, and now both sides have a total charge of 2-. But we have an extra O and two extra H's on the left. That is easy to fix. Just add water.
Cl2 (g) + 2 OH- (aq) ---> Cl- (aq) + ClO- (aq) + H2O (l)