Calculate the standard enthalpy of reaction:
2F2 (?) + 2H2O(?) → 4HF(?) + O2 (?).
∆?°? (F2, ?) = 0 kJ mol
∆?°? (O2, ?) = 0 kJ/mol
∆?°? (HF,?) = −268.61 kJ/mol
∆?°? (H2O, ?) = −285.8 kJ/mol
Calculate the standard enthalpy of reaction: 2F2 (?) + 2H2O(?) → 4HF(?) + O2 (?). ∆?°?...
A 70-g small metal ball was dropped in a 150 mL water contained in a constant-pressure calorimeter. The changed in water temperature is observed to be 1.21C°. Calculate the heat lost by the small metal ball. Calculate the standard enthalpy of reaction: 2F2(?)+2H2O(?)→4HF(?)+O2(?). Δ?°? (F2,?)=0kJmol; Δ?°? (O2,?)=0 kJ/mol; Δ?°? (HF,?)=−268.61 kJ/mol; Δ?°? (H2O,?)=−285.8 kJ/mol
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
(), Calculate the standard enthalpy change for the reaction Sio, (s) + 4HF (g) → SiF, (g) + 2H, given the information in the table below. Compound .H (kJ/mol) HF (g) -273.0 HO (1) -285.8 SiF, (g) -1614.9 Sio, (s) -910.9 Click the answer you think is right. -4189.4 kJ -2005.4 kJ +183.6 kJ -183.6 kJ
calculate enthalpy of H for the reaction N2H4(l) + 2H2O(l) -> N2(g) + 4H2)(l) Given the reactions N2H4(l) + O2(g) -> N2(g) + 2H2O(l) Enthalpy of H = -6.22.2 kJ H2(g) + (1/2)O2(g) -> H2O(l) enthalpy of H = -285.8 kJ/mol H2(g) + O2(g) -> H2O2(l) enthalpy of H = -187.8 kJ
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
Hess's Law Practice Find AH° for the following equation: SiO2(s) + 4HF(g) → SiF4(g) + 2H2O(g) Using the following equations: Si(s) + O2(g) → SiO2(s) Si(s) + 2F2(g) → SiF4(g) H2(g) + F2(g) → 2HF(g) H2(g) + 4202(g) → H2O(g) AH = -910.9 kJ/mol rxn AH = -1651 kJ/mol rxn AH = -542 kJ/mol rxn AH = -241.8 kJ/mol rxn
The change in Hrxn for the reaction C8H8 + 10 O2 = 8 CO2 + 4 H2O is -4395 kJ/mol. Knowing this, as well as the standard enthalpies of formation below, calculate the standard enthalpy of formation of Styrene in kJ/mol. change in Hf[CO2 (g)] = -393.5 kJ/mol change in Hf[H2O (l)] = -285.8 kJ/mol
A.Using standard heats of formation, calculate the standard enthalpy change for the following reaction. N2(g) + 3H2(g) = 2NH3(g) B.Using standard heats of formation, calculate the standard enthalpy change for the following reaction. CaCO3(s) = CaO(s) + CO2(g) C. A scientist measures the standard enthalpy change for the following reaction to be -2910.6 kJ: 2C2H6(g) + 7 O2(g) = 4CO2(g) + 6 H2O(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy...
Calculate the standard entropy change for the reaction 2H2(g)+O2(g)?2H2O(l) using the data from the following table: Substance ?H?f (kJ/mol) ?G?f (kJ/mol) S? [J/(K?mol)] H2(g) 0.00 0.00 130.6 O2(g) 0.00 0.00 205.0 H2O(l) -285.8 -237.2 69.90 Express your answer to four significant figures and include the appropriate units. Please show me the steps on how to solve this!!! Thank you!!!
Calculate the standard enthalpy of reaction for the combustion of propane. NOTE: This equation is not balanced. Round to the nearest whole number. C3H8(g) + O2 --> CO2(g) + H2O(l) kJ/mol Compound Hf (kJ/mole) C3H8(g) -105 CO2(g) -394 H2O(l) -284