Applying Hess’s Law, from the enthalpies of reactions,
2NaCl(s) + H2O(l) --> 2HCl(g) + Na2O(s) ΔH = + 507.31 kJ
NO(g) + NO2(g) + Na2O(s) --> 2NaNO2(s) ΔH = − 427.14 kJ
NO(g) + NO2(g) --> N2O(g) + O2(g) ΔH = − 42.68 kJ
2HNO2(l) --> N2O(g) + O2(g) + H2O(l) ΔH = + 34.35 kJ
Calculate the enthalpy change (ΔHrxn) for the reaction:
HCl(g) + NaNO2(s) --> HNO2(l) + NaCl(s) (You should show work to get credit)
5-Magnesium burns in air to produce a
bright light and is often used in fireworks displays.
The
combustion of magnesium produces magnesium
oxide solid. This reaction releases
1203 kJ of energy.
3-Write a balanced thermochemical equation for the combustion of
magnesium.
(b) How much heat (in kJ) is released by the combustion of 8.33 g
of magnesium?
4. Applying Hess’s Law, from the enthalpies of reactions,
Applying Hess’s Law, from the enthalpies of reactions, 2NaCl(s) + H2O(l) --> 2HCl(g) + Na2O(s) ΔH...
Help me get the answer -78.5 2) Use Hess's law to calculate the enthalpy of the reaction below using the following information. [Ans: -78.5] HCl(g) + NaNO2(s) → HNO2(1) + NaCl(s) AH=? 2NaCl(s) + H2O(l) + 2HCl(g) + Na2O(s) NO(g) + NO2(g) + Na2O(s) - 2NaNO3(s) NO(g) + NO2(g) N2O(g) + O2(g) 2HNO2(1) - N2O(g) + O2(g) + H2O(1) AH = 507 kJ AH = -427 kJ AH = -43 kJ AH = 34 kJ
Use the enthalpies for the given reactions to compute the standard enthalpy change (ΔH°) for the reaction: 6 C (s) + 7 H2(g) → C6H14 (l) ΔHrxn = ? C6H14 (l)+19/2O2 (g)→6CO2(g)+ 7H2O(g). ΔH=–3505.8kJ C(s) + O2 (g) → CO2(g) ΔH=–393.5kJ H2(g) + 1/2 O2(g) → H2O(g). ΔH = – 242.0 kJ
Using Hess’s Law to Calculate ΔH Calculate ΔH for 2 NO(g) + O2(g) → N2O4(g) using the following information: N2O4(g)2 NO(g) + O2(g)→→2 NO2(g)2 NO2(g)ΔHΔH==+57.9 kJ−114.1 kJ Calculate for using the following information: Select one 2.7 kJ -55.2 kJ -85.5 kJ -171.0 kJ +55.2 kJ
Use Hess' Law to calculate the ΔH for the following reaction from the data given below. Na2CO3(s) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l) ΔH = _________ Na2CO3(s) → CO2(g) + Na2O(s) ΔH = 319.8 kJ 2NaCl(aq) + H2O(l) → 2HCl(aq) + Na2O(s) ΔH = +348.0 kJ
(6 pts.) Using Hess’s Law and the values for Standard Enthalpies of Formation from the table provided, calculate the enthalpy of reaction ΔH°rxn (in kJ) for each of the following reactions: Standard Enthalpies of Formation substance ΔHf° in kJ/mol Mg(s) 0 MgO(s) -601.6 HCl(aq) -167.2 MgCl2(aq) -801.2 H2(g) 0 H2O(l) -285.8 Reaction #1: Mg (s) + 2HCl (aq) è MgCl2 (aq) + H2 (g) Reaction #2: MgO (s) + 2HCl (aq) è MgCl2 (aq) + H2O (l) 2. (4 pts.)...
Use Hess’s law to calculate ∆H° for the reaction:C(s) + 2H2(g) + ½O2(g) → CH3OH(l) ∆H°∘= ?using only the following data:H2(g) + ½O2(g) → H2O(l) ∆H°= -285.8 kJC(s) + O2(g) → CO2(g) ∆H°= -393.5 kJ2CH3OH(g) + 3O2(g) → 2CO2(g) + 4H2O(l) ∆H°= -1452.8 kJ
Part A Calculate the enthalpy of the reaction 2NO(g)+O2(g)→2NO2(g) given the following reactions and enthalpies of formation: 12N2(g)+O2(g)→NO2(g), ΔH∘A=33.2 kJ 12N2(g)+12O2(g)→NO(g), ΔH∘B=90.2 kJ Express your answer with the appropriate units. Part B Calculate the enthalpy of the reaction 4B(s)+3O2(g)→2B2O3(s) given the following pertinent information: B2O3(s)+3H2O(g)→3O2(g)+B2H6(g), ΔH∘A=+2035 kJ 2B(s)+3H2(g)→B2H6(g), ΔH∘B=+36 kJ H2(g)+12O2(g)→H2O(l), ΔH∘C=−285 kJ H2O(l)→H2O(g), ΔH∘D=+44 kJ Express your answer with the appropriate units.
Calculate ΔHrxn for the following reaction: 5C(s)+6H2(g)→C5H12(l) Use the following reactions and given ΔH′s. C5H12(l)+8O2(g)→5CO2(g)+6H2O(g), ΔH= -3244.8 kJ C(s)+O2(g)→CO2(g), ΔH= -393.5 kJ 2H2(g)+O2(g)→2H2O(g), ΔH= -483.5 kJ Express your answer to for significant figures.
Calculate ΔHrxn for the following reaction: 5C(s)+6H2(g)→C5H12(l) Use the following reactions and given ΔH values: C5H12(l)+8O2(g)→5CO2(g)+6H2O(g), ΔH= -3244.8 kJ C(s)+O2(g)→CO2(g), ΔH= -393.5 kJ 2H2(g)+O2(g)→2H2O(g), ΔH= -483.5 kJ
Consider the following chemical reaction. NH3(g) + 2 O2(g) → HNO3(aq) + H2O(l) Calculate the change in enthalpy (ΔH) for this reaction, using Hess' law and the enthalpy changes for the reactions given below. (1a) 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(l); ΔH = −1166.0 kJ/mol (2a) 2 NO(g) + O2(g) → 2 NO2(g); ΔH = −116.2 kJ/mol (3a) 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g); ΔH = −137.3 kJ/mol