Consider the following three reactions: (i) Ti(s) + 2 Cl 2 (g) → TiCl 4 (1g) (a) For each of the reactions, use data in Appendix C to calculate ΔH°, ΔG°, K, and ΔS ° at 25 °C.

Hello everyone. So in this video we want to sell for our delta H. Delta G. Are constant. And our delta S. Value for this reaction over here. And of course we're given all the necessary values that we will be using to sell for each and this table. So first things first we can go ahead and solve for our adult h delta ages ego to the delta age of our products minus the delta age of our attractants. So taking a look at my reaction, then we have the S. N. C. L. Four that being my product and these two the essence and chlorine is going to be my reactant. And we have given each term here. So let's go ahead and start. So my delta H. Let's see delta age of products. We will have negative 471.5 minus, let's see zero plus two times zero. That would give us the value of -471. killer jewels. So the reason why we have to multiply by two is because we have two moles. All right now we have the delta H. Wrong color here. We have delta H. Equaling two negative 4 71.5 kg jewels. So that's one of our answers here for this problem. Now moving on to our delta H. Or delta S. So same exact pattern. We have the values of our products minus the the values for our reactant. So my delta S. Is that equal to, let's see here we have 365.8 minus 51.2 times two times the 2 to 3.1 value. You put this into the calculator we get the value of -131. units being jewels per kelvin. So my delta S value is equal to negative 1 31.6 jewels per kelvin. So that's again one of my answers for this problem. Now solving for our delta G value again. Same exact pardon the delta G of my products minus the delta G. Of the reactant. So delta G. N. Is equal to. Let's see here we have 432.2 and that's negative plus let's see, zero plus two times zero. Putting that into the calculator, we get the value of negative 4 32 points to kill a jules. So our delta G value for this equation is equal to negative 432.2 kg jewels. So this is my third answer for this problem. I'll go ahead and scroll down just a little bit. We're gonna go ahead and squeeze in the main part or the last part of this um problem. So we're trying to solve now for our cake constant. So the reaction that we will or the equation that we're going to use for that is that delta G is equal to the negative R. Of T. Multiplied by the natural log of our k constant. We're isolating R k here. So we do so with some mathematical manipulation we get that are constant is equal to E raised to the power of delta G over negative R. Of T. So we can go ahead and plug in all the values then because we have we have calculated already for all these values. So the delta G. We said to be equal to negative 432. R. Value for delta G that we saw for is in kilo jewels. We needed to be in jewels. So we can go ahead and multiply this by and units being drooled. Then R. R. Is a constant. That is equal to 8.314 jewels. Polar molt times kelvin and their T value. Let's see, so T. Is given to us in Celsius in the problem. So to convert this into kelvin's, we have 25 plus. Let's see here To 73.5 or 15. And that would give us The Kelvin's 2-98.15. So that's the value. Can go ahead and plug in then to 98.15 Kelvin's. So once I put this big fraction into my calculator, I will get the value of 174.4. No Putting again this value into my calculator, I'll get the key constant being 5.2769 times to the 75. So then round it to the correct number of significant figures. We get that K. Is equal to five times 10 to the 75 power. So this is going to be my last answer for this problem. So again, we have our delta H, delta, S, delta G, and r K constant.

Ch.19 - Chemical Thermodynamics

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Brown 14th Edition

Ch.19 - Chemical Thermodynamics

Problem 95a

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Chapter 19, Problem 95a

Consider the following three reactions: (i) Ti(s) + 2 Cl₂(g) → TiCl₄(1g) (a) For each of the reactions, use data in Appendix C to calculate ΔH°, ΔG°, K, and ΔS ° at 25 °C.

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Identify the standard enthalpies of formation (ΔHf°) for each reactant and product from Appendix C. For the reaction Ti(s) + 2 Cl2(g) → TiCl4(l), find the ΔHf° values for Ti(s), Cl2(g), and TiCl4(l).

Calculate the standard enthalpy change (ΔH°) for the reaction using the formula: ΔH° = ΣΔHf°(products) - ΣΔHf°(reactants). Plug in the values you found for each substance.

Identify the standard Gibbs free energy of formation (ΔGf°) for each reactant and product from Appendix C. Use these values to calculate the standard Gibbs free energy change (ΔG°) for the reaction using the formula: ΔG° = ΣΔGf°(products) - ΣΔGf°(reactants).

Calculate the standard entropy change (ΔS°) for the reaction using the formula: ΔS° = ΣS°(products) - ΣS°(reactants). Use the standard molar entropy values (S°) from Appendix C for each reactant and product.

Determine the equilibrium constant (K) at 25 °C using the relationship between ΔG° and K, which is given by the equation: ΔG° = -RT ln(K), where R is the gas constant (8.314 J/mol·K) and T is the temperature in Kelvin (298 K for 25 °C). Solve for K.

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Key Concepts

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

Thermodynamics

Thermodynamics is the study of energy transformations and the relationships between heat, work, and energy. In chemical reactions, it helps determine the changes in enthalpy (ΔH°), Gibbs free energy (ΔG°), and entropy (ΔS°). Understanding these concepts is crucial for predicting the spontaneity and equilibrium of reactions.

Gibbs Free Energy

Gibbs free energy (G) is a thermodynamic potential that measures the maximum reversible work obtainable from a thermodynamic system at constant temperature and pressure. The change in Gibbs free energy (ΔG°) indicates whether a reaction is spontaneous (ΔG° < 0) or non-spontaneous (ΔG° > 0) under standard conditions.

Equilibrium Constant (K)

The equilibrium constant (K) is a dimensionless value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given reaction at a specific temperature. It is related to the Gibbs free energy change (ΔG°) by the equation ΔG° = -RT ln(K), where R is the gas constant and T is the temperature in Kelvin.

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Equilibrium Constant K

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Textbook Question

A standard air conditioner involves a refrigerant that is typically now a fluorinated hydrocarbon, such as CH₂F₂. An air-conditioner refrigerant has the property that it readily vaporizes at atmospheric pressure and is easily compressed to its liquid phase under increased pressure. The operation of an air conditioner can be thought of as a closed system made up of the refrigerant going through the two stages shown here (the air circulation is not shown in this diagram).

During expansion, the liquid refrigerant is released into an expansion chamber at low pressure, where it vaporizes. The vapor then undergoes compression at high pressure back to its liquid phase in a compression chamber. (e) Suppose that a house and its exterior are both initially at 31 °C. Some time after the air conditioner is turned on, the house is cooled to 24 °C. Is this process spontaneous or nonspontaneous?

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Textbook Question

Trouton’s rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporization is about 88 J/mol‐K. b. Look up the normal boiling point of Br₂ in a chemistry handbook or at the WebElements website (www.webelements.com) and compare it to your calculation. What are the possible sources of error, or incorrect assumptions, in the calculation?

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Textbook Question

(c) In general, under which condition is ΔG°_f more positive (less negative) than ΔH°_f ? (i) When the temperature is high, (ii) when the reaction is reversible, (iii) when ΔS°_f is negative.

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Consider the following three reactions: (i) Ti(s) + 2 Cl₂(g) → TiCl₄(1g) (ii) C₂H₆(g) + 7 Cl₂(g) → 2 CCl₄(g) + 6 HCl(g) (iii) BaO(s) + CO₂(g) → BaCO₃(s) (c) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature.

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Using the data in Appendix C and given the pressures listed, calculate K_p and ΔG for each of the following reactions: (c) N₂H₄(g) → N₂(g) + 2 H₂(g) P_N2H4 = 0.5 atm, P_N2 = 1.5 atm, P_H2 = 2.5 atm

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(a) For each of the following reactions, predict the sign of ΔH° and ΔS° without doing any calculations. (i) 2 Mg(s) + O₂ (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I₂(g) (iii) Na₂(g) ⇌ 2 Na(g) (iv) 2 V₂O₅(s) ⇌ 4 V(s) + 5 O₂(g)

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Consider the following three reactions: (i) Ti(s) + 2 Cl2(g) → TiCl4(1g) (a) For each of the reactions, use data in Appendix C to calculate ΔH°, ΔG°, K, and ΔS ° at 25 °C.

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Key Concepts

Thermodynamics

Gibbs Free Energy

Equilibrium Constant (K)

Consider the following three reactions: (i) Ti(s) + 2 Cl₂(g) → TiCl₄(1g) (a) For each of the reactions, use data in Appendix C to calculate ΔH°, ΔG°, K, and ΔS ° at 25 °C.