Methanol (CH 3 OH) can be made by the reaction of CO with H 2 : CO(𝑔) + 2 H 2 (𝑔) ⇌ CH 3 OH(𝑔) (a) Use thermochemical data in Appendix C to calculate ΔH° for this reaction.

hey everyone today, we're being asked to find the standard entropy of the reaction for the reaction of phosphorus becoming di phosphorus and tetra phosphorous at equilibrium. So to do this, we need to utilize the standard entropy of formation in order to find the standard entropy of the reaction as a whole. And we can use a general formula that goes as such. The standard entropy of the reaction reaction is equal to the difference between the standard entropy of the formation of the products and the standard entropy of formation of the reactant reactant. So with that in mind we actually need to find the specific values for the standard entropy of formation for each of the compounds present in the reaction. So let's go ahead and take a look at phosphorus first. So phosphorus, the standard entropy of formation, which is a experimentally calculated value and can be found in tables either online or in a textbook, but for phosphorus specifically phosphorus gas, The entropy of formation is 316. Kill it jules per mole for di phosphorus down to h the standard entropy of formation is 144 killer jewels per mole. And for tetra phosphorous, a standard entropy of formation is 58 0.9 kg joules per mole. So putting this into our products products minus reaction formula and let's keep our products in blue. We also need to consider how many moles of everything we have. So for the products we only have one mole produced of each. So we have one mole times 144 Kill it, jules per mole plus uh, one more Times 58.9 kg per mole. So this will be the standard entropy of formation for the products. And from here we will subtract the standard entropy of formation of the reactant. And we have six moles of loan phosphorus, that's what we have right here. So we actually factor that in as well. So we have six moles of p Times the 316.5 kila jules per mole and adding it all up. We get a final value. The final standard entropy of the reaction of negative 1696.1 kg jewels. I hope this helps. And I look forward to seeing you all in the next one.

Ch.15 - Chemical Equilibrium

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Brown 15th Edition

Ch.15 - Chemical Equilibrium

Problem 66a

Chapter 15, Problem 66a

Methanol (CH₃OH) can be made by the reaction of CO with H₂: CO(𝑔) + 2 H₂(𝑔) ⇌ CH₃OH(𝑔) (a) Use thermochemical data in Appendix C to calculate ΔH° for this reaction.

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Identify the standard enthalpy of formation (ΔH_f°) for each compound involved in the reaction from Appendix C.

Write the balanced chemical equation for the reaction: CO(g) + 2 H_2(g) ⇌ CH_3OH(g).

Use the formula for the standard enthalpy change of the reaction: ΔH° = Σ(ΔH_f° of products) - Σ(ΔH_f° of reactants).

Substitute the ΔH_f° values for CH_3OH(g), CO(g), and H_2(g) into the formula. Remember that the ΔH_f° for elements in their standard state, like H_2(g), is zero.

Calculate the sum of the enthalpies for the products and the reactants separately, then find the difference to determine ΔH° for the reaction.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Thermochemical Data

Thermochemical data includes information about the enthalpy changes associated with chemical reactions. This data is often found in tables and can provide standard enthalpy of formation values (ΔH°f) for various substances. By using these values, one can calculate the overall enthalpy change (ΔH°) for a reaction by applying Hess's law, which states that the total enthalpy change is the sum of the enthalpy changes for individual steps.

Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the number of steps or the pathway taken. This principle allows chemists to calculate the enthalpy change of a reaction by summing the enthalpy changes of individual reactions that lead to the same products. It is particularly useful when direct measurement of ΔH° is difficult or impossible.

Standard Enthalpy of Formation (ΔH°f)

The standard enthalpy of formation (ΔH°f) is defined as the change in enthalpy when one mole of a compound is formed from its elements in their standard states. This value is crucial for calculating the enthalpy change of a reaction using the formula ΔH° = ΣΔH°f(products) - ΣΔH°f(reactants). Understanding how to use ΔH°f values allows for the determination of the energy changes involved in chemical reactions.