determine δg°rxn using the following information. feo(s) co(g) → fe(s) co2(g) δh°= -11.0 kj; δs°= -17.4 j/k

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Answer 1

The standard free energy of reaction (ΔG°rxn) is calculated from the enthalpy of reaction (ΔH°rxn) and entropy of reaction (ΔS°rxn) using the formula:ΔG°rxn = ΔH°rxn - TΔS°rxnwhere ΔH°rxn is the standard enthalpy of reaction, ΔS°rxn is the standard entropy of reaction, and T is the absolute temperature in kelvins (K).

Given the following information, determine ΔG°rxn for the reaction: FeO(s) + CO(g) → Fe(s) + CO2(g)ΔH°rxn = -11.0 kJΔS°rxn = -17.4 J/First, we need to convert ΔS°rxn from joules per kelvin to kilojoules per kelvin.ΔS°rxn = -17.4 J/K = -0.0174 kJ/KNow, we can substitute the values into the formula and solve for ΔG°rxn:ΔG°rxn = ΔH°rxn - TΔS°rxnΔG°rxn = (-11.0 kJ) - (298 K)(-0.0174 kJ/K)ΔG°rxn = -11.0 kJ + 5.19 kJΔG°rxn = -5.81 kJ Therefore, the standard free energy of reaction is ΔG°rxn = -5.81 kJ.

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Related Questions

which of the choice(s) shows a keto-enol tautomeric pair(s)?

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option B shows a keto-enol tautomeric pair(s).

Option A (2-propanol and propanone),

option C (n-butane and isobutane), and

option D (2-methyl-2-propanol and 2-butanol) do not show a keto-enol tautomeric pair(s).

A keto-enol tautomeric pair is a pair of isomers, where one molecule contains a keto group while the other molecule contains an enol group. Keto and enol forms are tautomers because they can easily interconvert. When an alpha-hydrogen is present adjacent to a carbonyl group, the keto-enol tautomerization process occurs spontaneously, and it is a reversible process. Therefore, option (B) Acetone and propen-2-ol shows a keto-enol tautomeric pair. Acetone is a ketone, and propen-2-ol is an enol. The following equilibrium is established between them;

Acetone ⇔ Propen-2-ol

Thus, option B shows a keto-enol tautomeric pair(s).

Option A (2-propanol and propanone),

option C (n-butane and isobutane), and

option D (2-methyl-2-propanol and 2-butanol) do not show a keto-enol tautomeric pair(s).

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a mixture of two gases with a total pressure of 2.02 atm contains 0.70 atm of gas a. what is the partial pressure of gas b in atm?

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Given that a mixture of two gases with a total pressure of 2.02 atm contains 0.70 atm of gas. Hence, the partial pressure of gas b is 1.32 atm.

a. We need to find the partial pressure of gas b in atm. Let the partial pressure of gas b be Pb given that: Total pressure of the mixture, P = 2.02 atm Partial pressure of gas a, Pa = 0.70 atm Partial pressure of gas b, Pb = ? From Dalton's law of partial pressures, the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases in the mixture. P = Pa + Pb Substitute the given values, Pb = P - Pa= 2.02 atm - 0.70 atm= 1.32 atm.

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write the chemical formula of the following complex ions. hexabromomanganate (iii)

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The chemical formula of the hexabromomanganate(III) complex ion is [MnBr₆]³-.The formula of a coordination complex is typically written as [Metal ligands].

To form the hexabromomanganate (III) complex, Mn(III) cation (Mn³⁺) reacts with six Br⁻ ions to form the [MnBr₆]³⁻ complex ion.The name of the complex indicates that the metal ion is manganese(III), with oxidation state of +3, and the ligand is bromide ion, Br⁻. The prefix "hexa-" indicates the number of ligands, which is six.

Therefore, the complex ion is hexabromomanganate(III).The formula for the complex ion can also be determined using the charge balance principle. Since each bromide ion carries a charge of -1, the total charge of the six bromide ions is -6. Therefore, the manganese ion must have a charge of +3 to balance the negative charge of the six bromide ions. Hence, the formula of the hexabromomanganate(III) complex ion is [MnBr₆]³-.This is the long answer to the question.

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