For the reaction
CH3OHsld 1 32
O2sgdh2H2Osld 1 CO2sgd
the value of ΔG ° is −702.2 kJ at 25°C. Other data are as
follows:
DHf 8 (kJ/mol) S 8(J/mol∙K) at 258C at 258C
CH3OH(l ) −238.7 126.8
H2O(l ) −285.8 70.0
CO2(g) −393.5 213.7
Calculate the standard entropy, S °, per mole of O2(g).
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For the reaction CH3OHsld 1 32 O2sgdh2H2Osld 1 CO2sgd the value of ΔG ° is −702.2...
1. 1) luciferin+O2 ⇌ oxyluciferin+light 2) ATP⇌AMP+PPi ΔG∘=−31.6 kJ/mol If the overall ΔG∘ of the coupled reaction is -1.21 kJ/mol , what is the equilibrium constant, K, of the first reaction at 11∘C? Round your answer to 3 significant figures. 2. When methanol (CH3OH) is combusted, such as when in a gasoline blend, the following reaction occurs: 2CH3OH(l)+3O2(g)→2CO2(g)+4H2O(g) Based on the standard free energies of formation given, what is the standard free energy change for this reaction? Report the answer...
The standard enthalpies of formation, at 25.00 oC, of methanol (CH4O(l)), water (H2O(l)), and carbon dioxide (CO2(g)) are, respectively, -238.7 kJ/mol, -285.8 kJ/mol, and -393.5 kJ/mol. Calculate the change in the entropy of the surroundings (in J/K) upon the combustion of 16.9 g of methanol under a constant pressure of 1.000 atm and a temperature of 25.00 oC. N.B. combustion is the reaction of this substance with molecular oxygen to produce water and carbon dioxide.
Calculate the enthalpy change for the reaction C (graphite) + 2 H2 (g) + ½ O2 (g) → CH3OH (ℓ) Using the following information: C (graphite) + O2 (g) → CO2 (g) ΔHf° = –393.5 kJ H2 (g) + ½ O2 (g) → H2O (ℓ) ΔHf° = –285.8 kJ CH3OH (ℓ) + 3/2 O2 (g) → CO2 (g) + 2 H2O (ℓ) ΔHrxn° = –726.4 kJ a. –238.7 kJ b. 1691.5 kJ c. –1691.5 kJ d. 47.1 kJ e. –47.1...
Please explain
Data: C(graphite) + O2(g) => CO2(g) AH = -393.5 kJ H2(g) + 1/2O2(g) => H2O(1) AH = -285.8 kJ CH3OH(1) + 3/202(9) A CO2(g) + 2H20(1) AH = -726.4 kJ Using the data above, calculate the enthalpy change for the reaction below. Reaction: C(graphite) + 2H2(g) + 1/2O2(g) => CH3OH(1) A. +238.7 kJ B.-238.7 kJ C. +548.3 kJ D.-548.3 kJ E. +904.5 kJ
Calculate the ΔG°rxn at 298 K using the following information. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -207.0 91.3 33.2 -285.8 S°(J/mol∙K 146.0 210.8 240.1 70.0
The combustion reaction of ethane is as follows. C2H6(g) + 7/2 O2(g) → 2 CO2(g) + 3 H2O(l) Using Hess's law and the reaction enthalpies given below, find the change in enthalpy for this reaction. reaction (1): C(s) + O2(g) → CO2(g) ΔH = −393.5 kJ/mol reaction (2): H2(g) + 1/2 O2(g) → H2O(l) ΔH = −285.8 kJ/mol reaction (3): 2 C(s) + 3 H2(g) → C2H6(g) ΔH = −84.0 kJ/mol
At 298 K, evaluate deltaG(kJ) and deltaE (V) for
2CH3OH(l) + 3O2(g) -> 4H2O(g) + 2CO2(g)
S (J/mol-K) AH°y (kJ/mol) So (J/mol-K) AH(kJ/mol) Substance Substance N2(g) CH-ОН() 126.8 0 191.5 -238.6 CO(g) NH3(g) 197.9 192.5 -110.5 -46.2 CO2(g) NO(g) 213.6 +90.4 210.6 -393.5 NO2(g) +33.8 240.5 H2(g) HNO3(aq) 130.6 146.0 -206.6 H2O( 69.9 -285.8 O2(g) 188.8 H2O(g) 0 205.0 -241.8
Calculate the ΔG°rxn using the following information. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) ΔG°rxn = ? ΔH°f (kJ/mol) -207.0 91.3 33.2 -285.8 S°(J/mol∙K 146.0 210.8 240.1 70.0 -151 kJ +50.8 kJ -186 kJ +222 kJ -85.5 kJ
Methanol, CH3OH (l), combusts according to the following equation: 2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l) ∆rHo (298 K) = −1452 kJ Here is a list of Entropies of formation: S (J K-1 mol-1) at 298 K CH3OH (l) =126.8 O2 (g) = 205.14 CO2 (g) = 213.74 H2O = (l) 69.91 (a) If the above reaction was used in a fuel cell, say, to perform work, what will be the maximum...
Calculate the entropy of the following reaction, using the table below. Enter your answer without units to one decimal. 2C6H6(g) + 15O2 (g) → 12CO2 (g) + 6H2O (g) Substance ΔHf° (kJ/mol) ΔGf° (kJ/mol) S° (J/mol∙K) CH3COOH (g) -484.3 -389.9 159.8 (CH3)2O (g) -184.1 -112.6 266.4 C2H6 (g) -84.0 -32.0 229.2 C2H5OH (g) -234.8 -167.9 281.6 C6H6 (g) 82.9 129.7 269.2 CO2 (g) -393.5 -394.4 213.8 H2O (g) -241.8 -228.6 188.8 H2O (l) -285.8 -237.1 70 O2 (g) 0.0 0.0...