/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 87 \(\mathrm{NiO}\) is to be reduce... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\mathrm{NiO}\) is to be reduced to nickel metal in an industrial process by use of the reaction $$ \mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g) $$ At \(1600 \mathrm{~K},\) the equilibrium constant for the reaction is \(K_{p}=6.0 \times 10^{2} .\) If a CO pressure of \(20 \mathrm{kPa}\) is to be employed in the furnace and total pressure never exceeds \(101.3 \mathrm{kPa}\), will reduction occur?

Short Answer

Expert verified
Reduction will not occur since the required pressure exceeds the maximum allowed pressure.

Step by step solution

01

Understand the Reaction and Equilibrium Condition

The provided reaction is \( \mathrm{NiO}(s) + \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s) + \mathrm{CO}_2(g) \). Since \( K_p = 6.0 \times 10^{2} \), which is the equilibrium constant in terms of partial pressures for the gases, this measures the tendency for the reaction to go to completion at equilibrium.
02

Write the Expression for \( K_p \)

\( K_p \) is defined as the ratio of the partial pressures of the products to the reactants, raised to the power of their stoichiometric coefficients. For the given reaction, \( K_p = \frac{P_{\mathrm{CO}_2}}{P_{\mathrm{CO}}} \) since \( \mathrm{NiO} \) and \( \mathrm{Ni} \) are solids and do not appear in the expression.
03

Identify the Total Pressure Requirement

It is given that the total pressure in the furnace should not exceed \( 101.3 \mathrm{kPa} \). The total pressure is the sum of the pressures of \( \mathrm{CO} \) and \( \mathrm{CO}_2 \), i.e., \( P_{\text{Total}} = P_{\mathrm{CO}} + P_{\mathrm{CO}_2} \).
04

Solve for \( P_{\mathrm{CO}_2} \) in Terms of Total Pressure

We express the unknown pressure as \( P_{\mathrm{CO}_2} = 101.3 - P_{\mathrm{CO}} \).
05

Use \( K_p \) to Solve for \( P_{\mathrm{CO}_2} \)

Substitute \( P_{\mathrm{CO}} = 20 \mathrm{kPa} \) and the expression for \( P_{\mathrm{CO}_2} \) into the \( K_p \) expression: \( 6.0 \times 10^{2} = \frac{P_{\mathrm{CO}_2}}{20} \). Solve for \( P_{\mathrm{CO}_2} \) to get \( P_{\mathrm{CO}_2} = 6.0 \times 10^{2} \times 20 = 12000 \mathrm{kPa} \).
06

Analyze the Feasibility of the Reaction Under Given Conditions

Since \( P_{\mathrm{CO}_2} = 12000 \mathrm{kPa} \) is much greater than the total allowable pressure of \( 101.3 \mathrm{kPa} \), the reaction cannot reach equilibrium under these conditions. Therefore, reduction will not occur with the given limits on total pressure.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equilibrium Constant in Chemical Reactions
Understanding the equilibrium constant, denoted as \( K_p \) for reactions involving gases, is crucial in predicting the direction and extent of chemical reactions. The equilibrium constant is a numerical value that expresses the ratio of the concentrations or pressures of products to reactants at equilibrium, each raised to the power of their respective stoichiometric coefficients. For the reaction \( \mathrm{NiO}(s) + \mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s) + \mathrm{CO}_2(g) \), the equilibrium constant is given as \( K_p = 6.0 \times 10^{2} \).
This specific \( K_p \) value indicates a strong tendency for the reaction to proceed toward the formation of products at equilibrium, thus favoring the right side of the equilibrium equation.
  • Calculating \( K_p \): For heterogeneous reactions like the one given, only gaseous components are considered in the expression. Thus \( K_p = \frac{P_{\mathrm{CO}_2}}{P_{\mathrm{CO}}} \).
  • Impact of \( K_p \) Value: A higher \( K_p \) (>1) suggests products are favored when the system reaches equilibrium, whereas a low \( K_p \) (<1) indicates reactants are preferred.
  • Practical Application: In industrial settings, understanding \( K_p \) helps in designing processes by determining whether conditions allow for a chemical reaction to reach completion or if adjustments in pressure or temperature are needed.
Chemical Equilibrium in Industrial Chemistry
In industrial chemistry, achieving and maintaining chemical equilibrium is key to efficient production processes. Industrial reactions are often carried out under controlled conditions to maximize yields and reduce costs.
Equilibrium represents a dynamic balance where the rate of the forward reaction equals the rate of the reverse reaction, leading to stable product and reactant concentrations.
  • Pressure and Temperature: These are crucial parameters in industrial processes. For instance, in the reduction of \( \mathrm{NiO} \) to \( \mathrm{Ni} \), high temperatures (1600 K) are necessary to ensure the reaction occurs at a feasible rate.
  • Total Pressure Limitations: In the context of the given problem, industrial reactors have pressure constraints. Exceeding these limits can be unsafe and economically impractical, affecting whether a reaction can actually reach equilibrium as required by \( K_p \).
  • Efficiency and Safety: Industrial processes must balance optimal reaction conditions with safety standards and operational costs. This includes ensuring that total pressure inside reactors does not surpass designated safe thresholds, as highlighted by the given total pressure limit of \( 101.3 \mathrm{kPa} \).
Role of Reduction Reactions
Reduction reactions play a significant role in transforming raw materials into usable products, notably in metallurgy, where metals are extracted from ores. In the context of the given reaction, nickel oxide is reduced to nickel metal.
Reduction reactions involve the gain of electrons or a decrease in oxidation state. In this case, carbon monoxide (\( \mathrm{CO} \)) acts as the reducing agent, transforming \( \mathrm{NiO} \) by donating electrons.
  • Importance of Reduction: The industrial reduction of metal oxides is essential for producing pure metals. This forms the backbone of industries such as automotive, aerospace, and construction.
  • Reducing Agents: Carbon-based compounds (like \( \mathrm{CO} \)) are common in these reactions because they are efficient at electron donation, thereby facilitating the extraction of metals from their oxides.
  • Industrial Implications: Processes must be designed with care to ensure effective reduction. Temperature and pressure specifics are critical to drive these reactions towards the formation of desired products efficiently.
Adhering to these principles allows industries to optimize output while maintaining environmental and economic sustainability.

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Most popular questions from this chapter

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 5.00-L flask at \(310 \mathrm{~K}\), \(40 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{P}\) for this reaction at \(310 \mathrm{~K}\). (c) According to Le Châtelier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture was transferred to a 1.00-L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when 2.00 mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 1.00-L vessel at \(310 \mathrm{~K}\).

Consider the reaction $$ \mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q) $$ At \(25^{\circ} \mathrm{C}\), the equilibrium constant is \(K_{c}=2.4 \times 10^{-5}\) for this reaction. (a) If excess \(\operatorname{CaSO}_{4}(s)\) is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4}\), what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}{ }^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{~L}\), what is the minimum mass of \(\operatorname{CaSO}_{4}(s)\) needed to achieve equilibrium?

For \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g), K_{p}=3.0 \times 10^{2}\) at \(700 \mathrm{~K}\). In a 2.00-L vessel, the equilibrium mixture contains \(1.17 \mathrm{~g}\) of \(\mathrm{SO}_{3}\) and \(0.105 \mathrm{~g}\) of \(\mathrm{O}_{2}\). How many grams of \(\mathrm{SO}_{2}\) are in the vessel?

When \(1.50 \mathrm{~mol} \mathrm{CO}_{2}\) and \(1.50 \mathrm{~mol} \mathrm{H}_{2}\) are placed in a 3.00-L container at \(395^{\circ} \mathrm{C}\), the following reaction occurs: \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\). If \(K_{c}=0.802,\) what are the concentrations of each substance in the equilibrium mixture?

A mixture of 0.140 mol of \(\mathrm{NO}, 0.060 \mathrm{~mol}\) of \(\mathrm{H}_{2}\), and 0.260 mol of \(\mathrm{H}_{2} \mathrm{O}\) is placed in a 2.0-L vessel at \(330 \mathrm{~K}\). Assume that the following equilibrium is established: $$ 2 \mathrm{NO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ At equilibrium \(\left[\mathrm{H}_{2}\right]=0.010 \mathrm{M} .(\mathbf{a})\) Calculate the equilibrium concentrations of \(\mathrm{NO}, \mathrm{N}_{2}\), and \(\mathrm{H}_{2} \mathrm{O} .\) (b) Calculate \(K_{c}\).

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