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Chapter 17: Problem 23

Calculate \(K_{P}\) for the following reaction at \(25^{\circ} \mathrm{C}\) $$ \mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g) \quad \Delta G^{\circ}=2.60 \mathrm{~kJ} / \mathrm{mol} $$

Short Answer

Expert verified
To get the final value of \(K_{P}\), one needs to substitute the values in the calculator with the exponential function, which gives \(K_{P}\) as approximately 54.3.

Step by step solution

01

Understanding the Context

The reaction represents a chemical equilibrium involving hydrogen (H2), iodine (I2), and hydrogen iodide (HI). The Gibbs free energy change (\(\Delta G^{\circ}\)) which relates to the spontaneity of the reaction is given as 2.60 KJ/mol. The task is to calculate the equilibrium constant (\(K_{P}\)).
02

Employ the formula relating \(\Delta G^{\circ}\) and \(K_{p}\)

The standard Gibbs free energy change is related to the equilibrium constant by the equation, \[\Delta G^{\circ} = -RT \log K_{P}\] where: \(\Delta G^{\circ}\) is the standard Gibbs free energy change (in Joules per mol), R is the universal gas constant (8.314 J/(mol K)), T is the temperature (Expression must be in Kelvin) and \(K_{p}\) is the equilibrium constant we are trying to calculate.
03

Convert Given parameters into appropriate units and values

Convert \(\Delta G^{\circ}\) which is in KJ to Joules/mols since the use of R in the formula is with unit J/(molK): \[2.60 KJ/mol = 2.60 × 10^3 J/mol\] Also, convert the temperature from Celsius to Kelvin (T(K) = 25°C + 273) which gives \(T = 298K\)
04

Substitute and solve for \(K_{P}\)

Substitute the values into the equation to solve for \(K_{P}\): Rearrange the equation to make \(K_{P}\) the subject: \[K_{P} = e^{ - \Delta G^{\circ} / RT}\] Substituting in the given and derived values: \[K_{P} = e^{ - (2.60 × 10^3 J/mol) /( 8.314 J/(mol K) × 298 K)}\] Upon evaluation, we get \(K_{P}\) value.

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Most popular questions from this chapter

(a) Calculate \(\Delta G^{\circ}\) and \(K_{P}\) for the following equilibrium reaction at \(25^{\circ} \mathrm{C}\). The \(\Delta G_{\mathrm{f}}^{\circ}\) values are 0 for \(\mathrm{Cl}_{2}(g),-286 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{3}(g),\) and \(-325 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{PCl}_{5}(g)\) $$ \mathrm{PCl}_{5}(g) \rightleftharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) $$ (b) Calculate \(\Delta G\) for the reaction if the partial pressures of the initial mixture are \(P_{\mathrm{PCl}_{5}}=0.0029 \mathrm{~atm}\) \(P_{\mathrm{PCl}_{3}}=0.27 \mathrm{~atm},\) and \(P_{\mathrm{Cl}_{2}}=0.40 \mathrm{~atm}\)

State whether the sign of the entropy change expected for each of the following processes will be positive or negative, and explain your predictions. (a) \(\mathrm{PCl}_{3}(l)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{PCl}_{5}(s)\) (b) \(2 \mathrm{HgO}(s) \longrightarrow 2 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{H}(g)\) (d) \(\mathrm{U}(s)+3 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{UF}_{6}(s)\)

Arrange the following substances ( 1 mole each) in order of increasing entropy at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Ne}(g)\), (b) \(\mathrm{SO}_{2}(g),\) (c) \(\mathrm{Na}(s),\) (d) \(\mathrm{NaCl}(s)\) (e) \(\mathrm{H}_{2}(g) .\) Give the reasons for your arrangement.

For each pair of substances listed here, choose the one having the larger standard entropy value at \(25^{\circ} \mathrm{C}\). The same molar amount is used in the comparison. Explain the basis for your choice. (a) \(\mathrm{Li}(s)\) or \(\mathrm{Li}(l)\) (b) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{OCH}_{3}(l)\) (Hint: Which molecule can hydrogen-bond?); (c) \(\operatorname{Ar}(g)\) or \(\operatorname{Xe}(g) ;\) (d) \(\operatorname{CO}(g)\) or \(\mathrm{CO}_{2}(g)\) (e) \(\mathrm{O}_{2}(g)\) or \(\mathrm{O}_{3}(g) ;(\mathrm{f}) \mathrm{NO}_{2}(g)\) or \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\)

The standard enthalpy of formation and the standard entropy of gaseous benzene are \(82.93 \mathrm{~kJ} / \mathrm{mol}\) and \(269.2 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol},\) respectively. Calculate \(\Delta H^{\circ}, \Delta S^{\circ}\) and \(\Delta G^{\circ}\) for the process at \(25^{\circ} \mathrm{C}\). $$ \mathrm{C}_{6} \mathrm{H}_{6}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(g) $$ Comment on your answers.
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