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Chapter 17: Problem 69

Consider the following reaction at \(800 . \mathrm{K}\) : $$\mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g)$$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{~atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{~atm}, P_{\mathrm{NF}_{3}}=0.48 \mathrm{~atm} .\) Calculate \(\Delta G^{\circ}\) for the reaction at \(800 . \mathrm{K}\).

Short Answer

Expert verified
The standard Gibbs free energy change for the reaction at 800 K is approximately -8665 J/mol.

Step by step solution

01

Calculate the Reaction Quotient (Q)

We will use the equilibrium partial pressures to calculate the reaction quotient Q. The equation for Q in terms of partial pressures is: \(Q = \frac{(P_{NF_3})^2}{P_{N_2} \times (P_{F_2})^3}\) Plug in the given partial pressures: \(Q = \frac{(0.48 \space atm)^2}{(0.021 \space atm) \times (0.063 \space atm)^3}\)
02

Calculate the Gibbs Free Energy Change (ΔG)

Now that we have the reaction quotient (Q), we can find the Gibbs free energy change (ΔG) using the relation: ΔG = -RT·lnQ Given temperature T = 800 K and using the gas constant R = 8.314 J/mol·K: ΔG = - (8.314 J/mol·K) × (800 K) × ln(Q) Solve for ΔG: ΔG = -15321.6 J/mol
03

Calculate the Standard Gibbs Free Energy Change (ΔG°)

Now, we can use the equation relating ΔG, ΔG°, and Q to solve for ΔG°. Rearrange the equation to isolate ΔG°: ΔG° = ΔG - RT·lnQ ΔG° = -15321.6 J/mol - (8.314 J/mol·K) × (800 K) × ln(Q) Plug in the given values and solve for ΔG°: ΔG° ≈ -8665 J/mol So, the standard Gibbs free energy change for the reaction at 800 K is approximately -8665 J/mol.

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Most popular questions from this chapter

Consider two reactions for the production of ethanol: $$\begin{array}{l}\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l) \\\ \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}(l)+\mathrm{H}_{2}(g) \end{array}$$ Which would be the more thermodynamically feasible at standard conditions? Why?

Consider the following reaction at \(25.0^{\circ} \mathrm{C}\) : $$2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)$$ The values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are \(-58.03 \mathrm{~kJ} / \mathrm{mol}\) and \(-176.6 \mathrm{~J} / \mathrm{K}\). mol, respectively. Calculate the value of \(K\) at \(25.0^{\circ} \mathrm{C}\). Assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are temperature independent, estimate the value of \(K\) at \(100.0^{\circ} \mathrm{C}\).

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ}\), for a reaction at \(25^{\circ} \mathrm{C}\). How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C}\) ? What assumptions are made?

The Ostwald process for the commercial production of nitric acid involves three steps: a. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}, \Delta G^{\circ}\), and \(K\) (at \(298 \mathrm{~K}\) ) for each of the three steps in the Ostwald process (see Appendix 4 ). b. Calculate the equilibrium constant for the first step at \(825^{\circ} \mathrm{C}\), assuming \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature. c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

Given the values of \(\Delta H\) and \(\Delta S\), which of the following changes will be spontaneous at constant \(T\) and \(P ?\) a. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=300 . \mathrm{K}\) b. \(\Delta H=+25 \mathrm{~kJ}, \Delta S=+100 . \mathrm{J} / \mathrm{K}, T=300 . \mathrm{K}\) c. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=+5.0 \mathrm{~J} / \mathrm{K}, T=298 \mathrm{~K}\) d. \(\Delta H=-10 . \mathrm{kJ}, \Delta S=-40 . \mathrm{J} / \mathrm{K}, T=200 . \mathrm{K}\)
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