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Chapter 18: Problem 30

The equilibrium constant \(K_{P}\) for the reaction $$ \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{COCl}_{2}(g) $$ is \(5.62 \times 10^{35}\) at \(25^{\circ} \mathrm{C}\). Calculate \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{COCl}_{2}\) at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
After calculating the expression, you would get that \(\Delta G_f^\circ \approx -289.78 \,kJ \cdot mol^{-1}\). This represents the Gibbs free energy change for the given reaction at \(25°C\).

Step by step solution

01

Convert Celsius to Kelvin

Firstly, we need to convert the temperature from Celsius to the Kelvin scale, because the formula for Gibbs free energy involves absolute temperature. It can be done by applying the formula, \(T(K) = T(°C) + 273.15\). Therefore, for \(T = 25°C\), in Kelvin it is \(298.15K\).
02

Identify Known Values

Next, recognize all known quantities from the exercise. We have the equilibrium constant \(K_P = 5.62 \times 10^{35}\), the universal gas constant \(R = 8.3145 \, JK^{-1}mol^{-1}\) or \(R = 0.0083145 \, kJ K^{-1} mol^{-1}\) for conciseness, and the temperature \(T = 298.15 K\) from the previous step.
03

Use the Formula

Now, use the relation \(\Delta G_f^\circ = -RT\ln K_P\) to calculate Gibbs free energy. Plug in the known values and compute the result.
04

Calculate the Gibbs Free Energy

By using the given equation and values, \(\Delta G_f^\circ = - (0.0083145 \, kJ K^{-1} mol^{-1}) \times (298.15 K) \times \ln(5.62 \times 10^{35})\), perform the calculation to obtain the value of \(\Delta G_f^\circ\).

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Most popular questions from this chapter

Consider the following Bronstead acid-base reaction at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{HF}(a q)+\mathrm{Cl}^{-}(a q) \rightleftharpoons \mathrm{HCl}(a q)+\mathrm{F}^{-}(a q) $$. (a) Predict whether \(K\) will be greater or smaller than unity, (b) Does \(\Delta S^{\circ}\) or \(\Delta H^{\circ}\) make a greater contribution to \(\Delta G^{\circ} ?\) (c) Is \(\Delta H^{\circ}\) likely to be positive or negative?

In the metabolism of glucose, the first step is the conversion of glucose to glucose 6 -phosphate: glucose \(+\mathrm{H}_{3} \mathrm{PO}_{4} \longrightarrow\) glucose 6 -phosphate \(+\mathrm{H}_{2} \mathrm{O}\) $$ \Delta G^{\circ}=13.4 \mathrm{~kJ} / \mathrm{mol} $$ Because \(\Delta G^{\circ}\) is positive, this reaction does not favor the formation of products. Show how this reaction can be made to proceed by coupling it with the hydrolysis of ATP. Write an equation for the coupled reaction and estimate the equilibrium constant for the coupled process.

Define free energy. What are its units?

Ammonium nitrate \(\left(\mathrm{NH}_{4} \mathrm{NO}_{3}\right)\) dissolves spontaneously and endothermically in water. What can you deduce about the sign of \(\Delta S\) for the solution process?

A student placed \(1 \mathrm{~g}\) of each of three compounds \(\mathrm{A}\) \(\mathrm{B},\) and \(\mathrm{C}\) in a container and found that after 1 week no change had occurred. Offer some possible explanations for the fact that no reactions took place. Assume that \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) are totally miscible liquids.
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