The equilibrium constant K for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus 1\(/ T\) for this reaction gives a straight line with a slope of $1.352 \times 10^{4} \mathrm{K}\( and a y-intercept of \)-14.51 .$ Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise \(85 .\)

Short Answer

Expert verified
The standard enthalpy change (∆H°) for the given reaction is approximately \(-1.123 \times 10^{5} \mathrm{J/mol}\) and the standard entropy change (∆S°) is approximately \(-120.7 \mathrm{J/mol\cdot K}\).

Step by step solution

01

Recall the van't Hoff equation

The van't Hoff equation relates the change in the equilibrium constant (K) with temperature (T), standard enthalpy change (∆H°) and the standard entropy change (∆S°), and is given by: \[ \ln(K)= \frac{-\Delta H^{\circ}}{R} \left( \frac{1}{T} \right) + \frac{\Delta S^{\circ}}{R} \] Where R is the ideal gas constant which is equal to 8.314 J/mol·K.
02

Identify the slope and y-intercept from the given data

We are given that the slope of the graph of ln(K) versus 1/T is \(1.352 \times 10^{4} \mathrm{K}\) and y-intercept is \(-14.51\). Since we have a straight line, we can compare the van't Hoff equation to the equation of a straight line, which is: \[ y = mx + c \] Here, y corresponds to ln(K), m corresponds to the slope, x corresponds to 1/T, and c corresponds to the y-intercept.
03

Find the value of ∆H°

Comparing the van't Hoff equation to the equation of a straight line, we can see that the slope, m, is given by: \[ m = \frac{-\Delta H^{\circ}}{R} \] Now we can solve for ∆H° using the given slope value: \[ -\Delta H^{\circ} = m \times R = (1.352 \times 10^{4} \mathrm{K}) \times (8.314 \mathrm{J/mol\cdot K}) \] \[ \Delta H^{\circ} = -1.123 \times 10^{5} \mathrm{J/mol} \]
04

Find the value of ∆S°

Similarly, we can find the value of ∆S° using the given y-intercept, c: \[ c = \frac{\Delta S^{\circ}}{R} \] Now we can solve for ∆S° using the y-intercept value: \[ \Delta S^{\circ} = c \times R = (-14.51) \times (8.314 \mathrm{J/mol\cdot K}) \] \[ \Delta S^{\circ} = -120.7 \mathrm{J/mol\cdot K} \]
05

Conclusion

After solving for ∆H° and ∆S°, we can conclude that the standard enthalpy change (∆H°) for the given reaction is approximately \(-1.123 \times 10^{5} \mathrm{J/mol}\) and the standard entropy change (∆S°) is approximately \(-120.7 \mathrm{J/mol\cdot K}\).

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Most popular questions from this chapter

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. $\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$ b. $2 \mathrm{CH}_{3} \mathrm{OH}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(g)$ c. \(\mathrm{HCl}(g) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{Cl}^{-}(a q)\)

Consider the system $$\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)$$ at \(25^{\circ} \mathrm{C}\) a. Assuming that \(G_{\mathrm{A}}^{\circ}=8996 \mathrm{J} / \mathrm{mol}\) and \(G_{\mathrm{B}}^{\circ}=11,718 \mathrm{J} / \mathrm{mol},\) calculate the value of the equilibrium constant for this reaction. b. Calculate the equilibrium pressures that result if 1.00 mole of \(\mathrm{A}(g)\) at 1.00 atm and 1.00 mole of \(\mathrm{B}(g)\) at 1.00 atm are mixed at \(25^{\circ} \mathrm{C} .\) c. Show by calculations that \(\Delta G=0\) at equilibrium.

Calculate \(\Delta S_{\text { surr }}\) for the following reactions at \(25^{\circ} \mathrm{C}\) and 1 \(\mathrm{atm}\) . a. $\mathrm{C}_{3} \mathrm{H}_{8}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 3 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)\Delta H^{\circ}=-2221 \mathrm{kJ}$ b. $2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \qquad \Delta H^{\rho}=112 \mathrm{kJ}$

Consider the following reaction: $\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \qquad K_{298}=0.090\( For \)\mathrm{Cl}_{2} \mathrm{O}(g)$ $$\Delta G_{\mathrm{f}}^{\circ}=97.9 \mathrm{kJ} / \mathrm{mol}$$ $$\Delta H_{\mathrm{f}}^{\circ}=80.3 \mathrm{kJ} / \mathrm{mol}$$ $$S^{\circ}=266.1 \mathrm{J} / \mathrm{K} \cdot \mathrm{mol}$$ a. Calculate \(\Delta G^{\circ}\) for the reaction using the equation $\Delta G^{\circ}=-R T \ln (K)$ b. Use bond energy values (Table 8.5\()\) to estimate \(\Delta H^{\circ}\) for the reaction. c. Use the results from parts a and b to estimate \(\Delta S^{\circ}\) for the reaction. d. Estimate \(\Delta H_{\mathrm{f}}^{\circ}\) and \(S^{\circ}\) for \(\mathrm{HOCl}(g)\) e. Estimate the value of \(K\) at \(500 . \mathrm{K}\) . f. Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) when $P_{\mathrm{H}_{2} \mathrm{O}}=18\( torr, \)P_{\mathrm{Cl}_{2} \mathrm{O}}=$ 2.0 torr, and \(P_{\mathrm{HOC}}=0.10\) torr.

True or false: High temperatures are favorable to a reaction both kinetically and thermodynamically. Explain.

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