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Chapter 20: Problem 68

The equilibrium constant \(\left(K_{P}\right)\) for the reaction $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ is \(4.0 \times 10^{-31}\) at \(25^{\circ} \mathrm{C}\) and \(2.6 \times 10^{-6}\) at \(1100^{\circ} \mathrm{C},\) the temperature of a running car's engine. Is this an endothermic or exothermic reaction?

Short Answer

Expert verified
This reaction is an endothermic reaction.

Step by step solution

01

Analyse the given equilibrium constants

We are given two equilibrium constants, \(K_{P}\) at \(25^{\circ} \mathrm{C}\) is \(4.0 \times 10^{-31}\) and at \(1100^{\circ} \mathrm{C}\), \(K_{P}\) is \(2.6 \times 10^{-6}\). The equilibrium constant \(K_{P}\) has increased as the temperature increased from \(25^{\circ} \mathrm{C}\) to \(1100^{\circ} \mathrm{C}\).
02

Deduce the type of reaction

Since the value of \(K_{P}\) has increased with an increase in temperature, the reaction favours the products at higher temperatures. Hence, the reaction is absorbing heat to move towards products. This is the characteristic of an endothermic reaction.

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Most popular questions from this chapter

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A person was found dead of carbon monoxide poisoning in a well-insulated cabin. Investigation showed that he had used a blackened bucket to heat water on a butane burner. The burner was found to function properly with no leakage. Explain, with an appropriate equation, the cause of his death.

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As stated in the chapter, carbon monoxide has a much higher affinity for hemoglobin than oxygen does. (a) Write the equilibrium constant expression \(\left(K_{\mathrm{c}}\right)\) for the following process: $$ \mathrm{CO}(g)+\mathrm{HbO}_{2}(a q) \rightleftharpoons \mathrm{O}_{2}(g)+\mathrm{HbCO}(a q) $$ where \(\mathrm{HbO}_{2}\) and \(\mathrm{HbCO}\) are oxygenated hemoglobin and carboxyhemoglobin, respectively. (b) The composition of a breath of air inhaled by a person smoking a cigarette is \(1.9 \times 10^{-6} \mathrm{~mol} / \mathrm{L} \mathrm{CO}\) and \(8.6 \times 10^{-3} \mathrm{~mol} / \mathrm{L} \mathrm{O}_{2} .\) Calculate the ratio of \([\mathrm{HbCO}]\) to \(\left[\mathrm{HbO}_{2}\right]\), given that \(K_{\mathrm{c}}\) is 212 at \(37^{\circ} \mathrm{C}\).

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