For the following reaction:
IBr(g) + 4F2(g) → IF5(g) +
BrF3(g)
Compound | ΔH°f (kJ mol-1) | S° (J mol-1 K-1) | ||
IBr (g) | 40.88 | 258.95 | ||
F2 (g) | 0.00 | 202.80 | ||
IF5 (g) | -840.31 | 334.50 | ||
BrF3 (g) | -255.59 | 292.30 |
Determine the temperature (to two decimal places in K) such that
the reaction is in equilibrium in its standard states.
Step 1:
Given:
Hof(IBr(g)) = 40.88 KJ/mol
Hof(F2(g)) = 0.0 KJ/mol
Hof(IF5(g)) = -840.31 KJ/mol
Hof(BrF3(g)) = -255.59 KJ/mol
Balanced chemical equation is:
IBr(g) + 4 F2(g) ---> IF5(g) + BrF3(g)
ΔHo rxn = 1*Hof(IF5(g)) + 1*Hof(BrF3(g)) - 1*Hof( IBr(g)) - 4*Hof(F2(g))
ΔHo rxn = 1*(-840.31) + 1*(-255.59) - 1*(40.88) - 4*(0.0)
ΔHo rxn = -1136.78 KJ
Step 2:
Given:
Sof(IBr(g)) = 258.95 J/mol.K
Sof(F2(g)) = 202.8 J/mol.K
Sof(IF5(g)) = 334.5 J/mol.K
Sof(BrF3(g)) = 292.3 J/mol.K
Balanced chemical equation is:
IBr(g) + 4 F2(g) ---> IF5(g) + BrF3(g)
ΔSo rxn = 1*Sof(IF5(g)) + 1*Sof(BrF3(g)) - 1*Sof( IBr(g)) - 4*Sof(F2(g))
ΔSo rxn = 1*(334.5) + 1*(292.3) - 1*(258.95) - 4*(202.8)
ΔSo rxn = -443.35 J/K
Step 3:
At equilibrium, ΔGo = 0.0 KJ/Mol
ΔGo = 0.0 KJ/mol
ΔHo = -1136.78 KJ/mol
ΔSo = -443.35 J/mol.K
= -0.44335 KJ/mol.K
use:
ΔGo = ΔHo - T*ΔSo
0.0 = -1136.78 - T *-0.4434
T = 2564.07 K
Answer: 2564.07 K
For the following reaction: IBr(g) + 4F2(g) → IF5(g) + BrF3(g) Compound ΔH°f (kJ mol-1)...
For the following reaction: 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l) Compound ΔH°f (kJ mol-1) S° (J mol-1 K-1) CH3OH (l) -238.40 127.19 O2 (g) 0.00 205.70 CO2 (g) -393.51 213.74 H2O (l) -285.83 69.91 Determine the temperature (to two decimal places in K) such that the reaction is in equilibrium in its standard states.
For the following reaction: 2CH4(g) + O2(g) → 2CO(g) + 4H2(g) Compound ΔH°f (kJ mol-1) S° (J mol-1 K-1) CH4 (g) -74.87 188.66 O2 (g) 0.00 205.70 CO (g) -110.53 197.66 H2 (g) 0.00 130.68 Calculate ΔG°rx (in kJ) at 345.31 K for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ). Assume ΔH°f and S° do not vary as a function of temperature.
For the following reaction: CS2(g) + 3O2(g) → CO2(g) + 2SO2(g) Compound ΔH°f (kJ mol-1) S° (J mol-1 K-1) CS2 (g) 116.70 237.80 O2 (g) 0.00 205.70 CO2 (g) -393.51 213.74 SO2 (g) -296.84 248.20 Calculate ΔG°rx (in kJ) at 794.8 K for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ). Assume ΔH°f and S° do not vary as a function of temperature.
For the following reaction: 2Na(s) + 2H2O(l) → 2NaOH(s) + H2(g) Compound ΔH°f (kJ mol-1) S° (J mol-1 K-1) Na (s) 0.00 51.30 H2O (l) -285.83 69.91 NaOH (s) -425.93 64.46 H2 (g) 0.00 130.68 Calculate ΔG°rx (in kJ) at 391.96 K for this reaction. Assume ΔH°f and S° do not vary as a function of temperature.
ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g) is ________ kJ, give the data below. IF (g) + F2 (g) → IF3 (g) ΔH = -390 kJ IF (g) + 2F2 (g) → IF5 (g) ΔH = -745 kJ
For the following reaction: 2C3H18(1) + 2502(g) → 16C02(g) + 18H20(1) Compound AH°F (kJ moll) Sº (3 mol-1 K-1) C8H18 ( 259.30 328.00 O2 (g) 0.00 205.70 CO2 (9) -393.51 213.74 H20 (1) -285.83 69.91 Determine the temperature (to two decimal places in K) such that the reaction is in equilibrium in its standard states.
For the reaction 2 BrF3(g) - Br2(g) + 3 F2(g) AG° = 470.3 kJ and AH° = 542.1 kJ at 267 K and 1 atm. This reaction is (reactant, product) favored under standard conditions at 267 K. The entropy change for the reaction of 2.38 moles of BrF3(g) at this temperature would be J/K.
Consider the Haber synthesis of gaseous NH3 (ΔH∘f = -46.1 kJ/mol; ΔG∘f = -16.5 kJ/mol): N2(g)+3H2(g)→2NH3(g) What are the equilibrium constants Kp and Kc for the reaction at 350 K ? You may assume that ΔH∘ and ΔS∘ are independent of temperature.
The standard enthalpy of formation (ΔH∘f) is the enthalpy change that occurs when exactly 1 mol of a compound is formed from its constituent elements under standard conditions. The standard conditions are 1 atm pressure, a temperature of 25 ∘C , and all the species present at a concentration of 1 M . A "standard enthalpies of formation table" containing ΔH∘f values might look something like this: Substance ΔH∘f H(g) 218 kJ/mol H2(g) 0 kJ/mol Ba(s) 0 kJ/mol Ba2+(aq) −538.4...
ΔH f f ° = -36.40 kJ/mol and S° = 245.91 J/mol K for CH 3 3 Br, determine the Gibbs free energy in kJ from the ΔH f f ° and S° values for the reaction CH 4 4 + Br 2 2 → CH 3 3 Br + HBr at 298 K. Give only the numerical part of your answer; i.e., don’t include units in your response.