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
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Use the enthalpies for the given reactions to compute the standard enthalpy change (ΔH°) for the...
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...
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.
A scientist measures the standard enthalpy change for the following reaction to be 2752.8 kJ : 6CO2(g) + 6H2O(1) C6H12O6 + 6 O2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of CH1:05 is kJ/mol A scientist measures the standard enthalpy change for the following reaction to be 53.9 LJ: CO2(g) + H2(g)— CO(g) + H2O() Based on this value and the standard enthalpies of formation for the other...
Calculate the change in enthalpy (ΔH) for the heat of formation of ethane, C2H6, using Hess' Law and the following reactions:a) 2C(S) + 2O2(g) → 2CO2(g), ΔH = -188 kcalb) C2H6(g) + (7/2)O2(g) → 2CO2(g) + 3H2O(l), ΔH = -373 kcalc) H2(g) + (1/2)O2(g) → H2O(l), ΔH = -68.3 kcal
The standard heat of formation, ΔH∘f, is defined as the enthalpy change for the formation of one mole of substance from its constituent elements in their standard states. Thus, elements in their standard states have ΔH∘f=0. Heat of formation values can be used to calculate the enthalpy change of any reaction. Consider, for example, the reaction 2NO(g)+O2(g)⇌2NO2(g) with heat of formation values given by the following table: Substance ΔH∘f (kJ/mol) NO(g) 90.2 O2(g) 0 NO2(g) 33.2 Then the standard heat...
A scientist measures the standard enthalpy change for the following reaction to be 2853.6 kJ : 6CO2(g) + 6 H2O(l)C6H12O6 + 6 O2(g) Based on this value and the standard enthalpies of formation for the other substances, the standard enthalpy of formation of H2O(l) is ?kJ/mol.
Use Hess’s Law to find the standard enthalpy change for the reaction CO2(g) → C(s) + O2(g) using only the following information. Show all your work, including any equations you use to obtain your answer and showing clearly how you obtained that answer. (3 pts.) H2O(l) → H2(g) + 1/2O2(g) C2H6(g) → 2C(s)+ 3H2(g) 2CO2(g) + 3H2O(l) → C 2H6(g) + 7/2O2(g) ∆Ho (kJ) 643 kJ 190.6kJ 3511.1 kJ
a) Use Hess's law to calculate the enthalpy change for the reaction: 3C(s) + 4H2(g) + ½O2(g) → C3H8O(l) Given the following thermochemical equations: 2C3H8O(l) + 9O2(g) → 6CO2(g) + 8H2O(l) ΔH = -4042.6 kJ/mol C(s) + O2(g) → CO2(g) ΔH = -393.51 kJ/mol H2(g) + ½O2(g) → H2O(l) ΔH = -285.83 kJ/mol (in kJ/mol) A: -267.7 B: -302.6 C: -341.9 D: -386.3 E: -436.5 F: -493.3 G: -557.4 H: -629.9 b) Define if the following statement is an endothermic process or exothermic...
1).From the standard enthalpies of formation, calculate ΔH°rxn for the reaction C6H12(l) + 9O2(g) → 6CO2(g) + 6H2O(l) For C6H12(l), ΔH°f = –151.9 kJ/mol (5 points) Substance ∆H°f , kJ/mol C6H12(l) –151.9 O2(g) 0 H2O(l) –285.8 CO2(g) –393.5 2).Determine the amount of heat (in kJ) given off when 1.26 × 104 g of ammonia are produced according to the equation N2(g) + 3H2(g) → 2NH3(g) ΔH°= –92.6 kJ/mol Assume that the reaction takes place under standard conditions at 25oC.
Calculate the enthalpy of the following reaction: C (s) + 2 H2 (g) --> CH4 (g) Given: C (s) + O2 (g) --> CO2 ΔH = -393 kJ H2 + 1⁄2O2 --> H2O. ΔH = -286 kJ CH4 + 2O2 --> CO2 + 2H2O ΔH = -892 kJ