ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g) is ________ kJ, give...
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
Homework Answers
We need ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g)
Given
IF (g) + F2 (g) → IF3 (g) ΔH = -390 kJ - - - - - (1)
IF (g) + 2F2 (g) → IF5 (g) ΔH = -745 kJ - - - - - - (2)
To get the desired solution, reverse the second reaction and add that to the first reaction.
Hence the second reaction will be
IF5 (g) -> IF (g) + 2F2 (g) ΔH = +745 kJ - - - - - - (3)
Since we reversed the reaction ΔH will become positive
Now
Add (3) and (1)
IF5 (g) -> IF (g) + 2F2 (g) ΔH = +745 kJ
IF (g) + F2 (g) → IF3 (g) ΔH = -390 kJ
IF (g) and one F2 (g) will be cancelled on reactants and products side
Hence
IF5 (g) → IF3 (g) + F2 (g) ΔH = +745 KJ - 390 KJ = +355 KJ
Therefore
ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g) is +355 KJ
Add Answer to:
ΔH for the reaction IF5 (g) → IF3 (g) + F2 (g) is ________ kJ,
give...
Change in enthalpy for the reaction IF5(g) ------> IF3(g) + F2(g) is ______kJ, give the data...
Change in enthalpy for the reaction IF5(g) ------> IF3(g) + F2(g) is ______kJ, give the data below. IF(g) + F2(g) ---->IF3(g) change in enthalpy = -390kJ IF(g) + 2F2(g)------->IF5(g) change in enthalpy = -745kJ
ΔH for the reaction IF_{5}5(g) → IF_{3}3(g) + F_{2}2 is kJ, give the data below. IF(g)...
ΔH for the reaction IF_{5}5(g) → IF_{3}3(g) + F_{2}2 is kJ, give the data below. IF(g) + F_{2}2(g) → IF_{3}3(g) ΔH = -390 kJ IF(g) + 2F_{2}2(g) → IF_{5}5(g) ΔH = -745 kJ
AH = -390 kJ AH = -745 kJ 13. Given the following thermochemical equations: IF (g)...
AH = -390 kJ AH = -745 kJ 13. Given the following thermochemical equations: IF (g) + F2 (g) F3 (9) IF (9) + 2 F2 (9) IFs (9) Find the value of AH for the equation below: (6 points) IFs (9) - IF3 (9) + F2 (9) 14. Baking soda (NaHCO3) can be decomposed when heated, as shown: 2 NaHCO3 (s) Na CO2 (s) + CO2(g) + H20 (g) Use heats of formation to determine the value of AH...
AH 20. HESS'S LAW. Calculate the AH for the reaction: IF.) → Fle) + Flel Use...
AH 20. HESS'S LAW. Calculate the AH for the reaction: IF.) → Fle) + Flel Use the following data: a. IF(g) + Fle) → 1f3(E) AH = -390 kJ b. IF(g) + 2F2(e) → Fle) AH = -745 +1135 - 1135
A chemist knows that the kJ for the reaction 2H2(g) + O2 (g) ---> 2H2O (g)...
A chemist knows that the
kJ for the reaction 2H2(g) + O2
(g) ---> 2H2O (g) ,and that
kJ
for the reaction H2 (g) + F2 (g) ---> 2HF
(g).
With this information he calculated the
for the reaction 2H2O (g) + 2F2
(g) ---> 4HF(g) + O2 (g) and
predicted whether
was positive or negative. How?
A Ho- 485
Given the following data: C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ CS2(l) + 3O2(g) → CO2(g) + 2SO2(g) ΔH = -1076.5 kJ C(s) +...
Given the following data: C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ CS2(l) + 3O2(g) → CO2(g) + 2SO2(g) ΔH = -1076.5 kJ C(s) + 2S(s) → CS2(l) ΔH = +89.4 kJ Find the ΔH of the following reaction: SO2(g) → S(s) + O2(g)
18. Given that AHorxn = -546.6 kJ H2 (g)+F2 (g)2HF (g) AH°rxn = -571.6 kJ 2H2...
18. Given that AHorxn = -546.6 kJ H2 (g)+F2 (g)2HF (g) AH°rxn = -571.6 kJ 2H2 (g)+ O2 (g) 2H20 (1) Calculate the value of AHrxn for 2F2 (g) 4HF (g) + O2 (g) +2H20 (I)
2. Use Hess’s Law to determine the enthalpy of the reaction below. 2F2(g) + 2H2O(l) → 4HF(aq) + O2(g) DH˚= ? H2(g) + F2(...
2. Use Hess’s Law to determine the enthalpy of the reaction below. 2F2(g) + 2H2O(l) → 4HF(aq) + O2(g) DH˚= ? H2(g) + F2(g) → 2HF(aq) DH˚ = -546.6 kJ 2H2 (g) + O2(g) → 2H2O(l) DH˚ = -571.6 kJ a. 42 kJ b. -1120 kJ c. -251 kJ d. -521 kJ e. -1690 kJ
At 25°C, the following heats of reaction are known: 2CIF(g) + O2(g) → Cl20(g) + F2。dH'm...
At 25°C, the following heats of reaction are known: 2CIF(g) + O2(g) → Cl20(g) + F2。dH'm = 167.4 kJ/mol 2ClF3(g) + 202(g) → Cl20(g) + 3F20(g) ΔH'm-34 1.4 kJ/mol 2F2(g) + O2(g) → 2F20(g) ΔHoxn--43.4 kJ/mol At the same temperature, use Hess's law to calculate ΔHorn for the reaction: CIF(g) + F2(g) → ClF3(g) 0 a-21 7.5 kJ/mol O b 465.4 kJ/mol O c. -108.7 kJ/mol d 217.5 kJ/mol O e-130.2 kJ/mol
Consider the following data. 2 H2(g) + O2(g) 2 H2O(l) ΔH = -571.7 kJ N2O5(g) +...
Consider the following data. 2 H2(g) + O2(g) 2 H2O(l) ΔH = -571.7 kJ N2O5(g) + H2O(l) 2 HNO3(l) ΔH = -92.0 kJ N2(g) + 3 O2(g) + H2(g) 2 HNO3O(l) ΔH = -348.2 kJ Use Hess's law to calculate ΔH for the reaction below. 2 N2O5(g) 2 N2(g) + 5 O2(g) ΔH = _____kJ
AH = -390 kJ AH = -745 kJ 13. Given the following thermochemical equations: IF (g) + F2 (g) F3 (9) IF (9) + 2 F2 (9) IFs (9) Find the value of AH for the equation below: (6 points) IFs (9) - IF3 (9) + F2 (9) 14. Baking soda (NaHCO3) can be decomposed when heated, as shown: 2 NaHCO3 (s) Na CO2 (s) + CO2(g) + H20 (g) Use heats of formation to determine the value of AH...
AH 20. HESS'S LAW. Calculate the AH for the reaction: IF.) → Fle) + Flel Use the following data: a. IF(g) + Fle) → 1f3(E) AH = -390 kJ b. IF(g) + 2F2(e) → Fle) AH = -745 +1135 - 1135
A chemist knows that the
kJ for the reaction 2H2(g) + O2
(g) ---> 2H2O (g) ,and that
kJ
for the reaction H2 (g) + F2 (g) ---> 2HF
(g).
With this information he calculated the
for the reaction 2H2O (g) + 2F2
(g) ---> 4HF(g) + O2 (g) and
predicted whether
was positive or negative. How?
A Ho- 485
18. Given that AHorxn = -546.6 kJ H2 (g)+F2 (g)2HF (g) AH°rxn = -571.6 kJ 2H2 (g)+ O2 (g) 2H20 (1) Calculate the value of AHrxn for 2F2 (g) 4HF (g) + O2 (g) +2H20 (I)
At 25°C, the following heats of reaction are known: 2CIF(g) + O2(g) → Cl20(g) + F2。dH'm = 167.4 kJ/mol 2ClF3(g) + 202(g) → Cl20(g) + 3F20(g) ΔH'm-34 1.4 kJ/mol 2F2(g) + O2(g) → 2F20(g) ΔHoxn--43.4 kJ/mol At the same temperature, use Hess's law to calculate ΔHorn for the reaction: CIF(g) + F2(g) → ClF3(g) 0 a-21 7.5 kJ/mol O b 465.4 kJ/mol O c. -108.7 kJ/mol d 217.5 kJ/mol O e-130.2 kJ/mol
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