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
Change in enthalpy for the reaction IF5(g) ------> IF3(g) + F2(g) is ______kJ, give the data...
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
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
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.
Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solid iodine and gaseous fluorine: I2(s)+5F2(g)→2IF5(g) A 5.50 −L flask containing 11.0 g I2 is charged with 11.0 g F2, and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 ∘C. A) What is the partial pressure of IF5 in the flask? B) What is the mole fraction of IF5 in the flask? C) Draw...
for the chemical reaction H2(g) + F2(g) → 2HF(g); ∆H°=-79.2 kj/mol what is the enthalpy for the reaction 3H2+3F2->6HF
What is the enthalpy of reaction, dH(rxn), for the following reaction? C2H4 (g). +. 6 F2 (g). --->. 2 CF4 (g). +.4 HF (g) dH(f) +52.3 -680 -268.5. kJ/mole dH(rxn) - +2486 kJ dH(rxn) = -2486 kJ o dH(rxn) = -2434 kJ • dH(rxn) = -1000.8 kJ
Calculate deltaH° fornthe following reaction: IF7(g) + I2(g) --> IF5(g) + 2IF(g) using the following information: IF5. -840 IF7. -941 IF. -95
Calculate the enthalpy change for the reaction NO(g) + O(g) → NO2(g) from the following data NO(g) + O3(9) → NO2(g) + O2(g) ΔH=-198.9 kJ/mol O3(g) → 1.5O26(g) ΔH=-142.3 kJ/mol O2(g) → 2O(g) ΔH = 495.0 kJ/mol A. 153.8 kJ B. 190.9 kJ C.-551.6 kJ D.-304.1 kJ E. 438.4 kJ
5) Calculate the enthalpy change for the reaction NO(g)+ Og) NO2g) from the following data: NOg) +03(g)-NO2(g) + O2(g) ΔΗ--I 98.9 kJ O3g) 1.502(g) 02(g) 20(g) AH-142.3 kJ AH = 495.0 kJ A) 153.8 kJ B) 438.4 kJ C)-551.6 kJ D) 190.9 kJ E)-304.1 kJ
TC04M01 Use the enthalpy changes given in the data below to calculate the enthalpy change for this reaction: Cus(s) + O2(g) → Cu(s) + SO2(9) Data: Reaction no. Reaction A.HⓇ/kJ morat 298K standard state = 1 bar. +314 2 2 CuO(s) – 2Cu(s) + O2(g) S(s) + O2(g) - S02(9) 2CuO(s) + 2S(s) - 2CuS(s) + O2(9) -297 +208 Select one: a. +225 kJ/mol b.-85 kJ/mol O C. -244 kJ/mol d. -191 kJ/mol e. -225 kJ/mol