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 process for the system? In each case, the “system” is underlined.
exothermic or endothermic water freezing into ice cubes
in the freezer
exothermic or endothermic dry ice subliming into carbon
dioxide
exothermic or endothermic a block of cheese being cooled
in a refrigerator
a) Use Hess's law to calculate the enthalpy change for the reaction: 3C(s) + 4H2(g) +...
Use the following information and hess's law to find the enthalpy change for the reaction C(g) + O2(g) => CO2(g) reaction 1: 2CO(g) + O2(g) => 2CO2(g) ΔH = -566 kj reaction 2: 2C(g) + O2(g) => 2CO(g) ΔH = -1655 kj
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.
21 Use Hess's Law to calculate the enthalpy change for the reaction: Wo3(s)+3H2(g)-W(s)+3H20(g) Eqn 1: 2W(s)302(g) 2WO3(s) AH -1685.4 kJ/mol Eqn 2: 2H2(g) O2(g)2H20(g) AH -477.84 kJ/mol Give your answer as just a number without the assumed units (kJ/mol)
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
Use Hess's law to determine Ofor the reaction → CO2(g),given that C(graphite) +202(8) — CO(8) A Hº = -110.54 kJ mol C(graphite) + O2(8) —CO2(g) A Hº = -393.51 kJ mol
Using Hess's Law, determine the enthalpy change of the following reaction. CS2(1) + 3 O2(g) --> CO2(g) + 2 SO2(g) C(s) + O2(g) --> CO2(g) AH = -393.5 kJ/mol S(s) + O2(g) --> SO2(g) AH = -296.8 kJ/mol C(s) + 25(s) --> CS2(8) AH = +87.9 kJ/mol -899.2 kJ O-778.2 kJ -602.4 kJ - 1075.0 kJ
Using Hess's Law, calculate ΔH°R Equation: B2H6 (g) + 6 Cl2 (g) --> 2BCl3 (g) + 6 HCl (g) Given these 3 equations: (please show all work) BCl3 (g) + 3H2O (l) --> H3BO3 (g) + 3HCl (g) (ΔH°R = -112.5 KJ/mol BCl3) B2H6 (g) + 6H2O (l) --> 2H3BO3 (g) + 6H2 (g) (ΔH°R = -493.4 KJ/mol B2H6) H2 (g) + Cl2 (g) --> 2HCl (g) (ΔH°R = -184.6 KJ/mol H2)
Calculate AH for the reaction 2 KCIO: (s) → 2 KCl (s) + 3 O2 (g) ΔH°r of KCIO3 (s) = -397.7 kJ/mol ΔH°r of KCI (s) = -436.5 kJ/mol
Calculate the standard enthalpy change for the reaction 2C8H18(l) + 17O2(g) →16CO(g) + 18H2O(l). Given 2C8H18(I) + 25O2(g) →16CO2(g) + 18H2O(I) ΔH°=-11,020 kJ/mol 2CO(g) + O2(g) → 2CO2(g) ΔH° = -566.0 kJ/mol 0 -6,492 kJ/mol 15,550 kJ/mol 10,450 kJ/mol -10.450 kJ/mol 6,492 kJ/mol
For the following reaction: 2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(l) Compound ΔH°f (kJ mol-1) S° (J mol-1 K-1) CH3OH (l) -238.40 127.19 O2 (g) 0.00 205.70 CO2 (g) -393.51 213.74 H2O (l) -285.83 69.91 Determine the temperature (to two decimal places in K) such that the reaction is in equilibrium in its standard states.