\(\mathrm{NiO}\) 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
Given that the equilibrium constant \(K_p = 6.0 \times 10^2\), the initial CO pressure is 20 kPa, and the initial CO2 pressure is 0 kPa, we can calculate the initial reaction quotient as \(Q = \frac{P_{CO_2}}{P_{CO}} = \frac{0}{20} = 0\). Since \(Q = 0 < K_p = 6.0 \times 10^2\), the reaction will proceed in the forward direction, leading to the reduction of NiO to Ni. Thus, the reduction will occur under the given conditions.

Step by step solution

01

Write the expression for the reaction quotient

The reaction quotient, Q, can be calculated by dividing the partial pressure of products by the partial pressure of reactants:$$Q = \frac{P_{CO_2}}{P_{CO}}$$ where \(P_{CO_2}\) is the partial pressure of CO2 and \(P_{CO}\) is the partial pressure of CO.
02

Determine the initial partial pressures of CO and CO2

We are given that the CO pressure to be employed in the furnace is 20 kPa, and the total pressure never exceeds 101.3 kPa. Since no partial pressure is given for CO2, we can assume that the initial partial pressure of CO2 is zero. Therefore, we have: $$P_{CO} = 20 \mathrm{kPa}$$$$P_{CO_2} = 0 \mathrm{kPa}$$
03

Calculate the initial reaction quotient, Q

Now, plug in the initial partial pressures of CO and CO2 that we found in step 2 into Q equation:$$Q = \frac{0 \mathrm{kPa}}{20 \mathrm{kPa}} = 0$$
04

Compare Q to Kp

We have calculated the Q value as 0, and the equilibrium constant Kp as \(6.0 \times 10^2\). Since $$Q = 0 < K_p = 6.0 \times 10^2$$ this means that the reaction will proceed forward, towards the products.
05

Determine if reduction occurs

Since Q < Kp, the reaction will proceed in the forward direction, which is the reduction of NiO to Ni. Therefore, the reduction of NiO will occur under the given conditions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate \(K_{c}\) at \(900 \mathrm{~K}\) for $2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\( if \)K_{p}=0.0572$ at this temperature.

At \(120^{\circ} \mathrm{C}, K_{c}=0.090\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2}\) are $0.100 \mathrm{M}\( and \)0.075 \mathrm{M}$, respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

At \(900^{\circ} \mathrm{C}, K_{c}=0.0108\) for the reaction $$\mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$$ A mixture of \(\mathrm{CaCO}_{3}, \mathrm{CaO},\) and \(\mathrm{CO}_{2}\) is placed in a \(10.0-\mathrm{L}\) vessel at \(900^{\circ} \mathrm{C}\). For the following mixtures, will the amount of \(\mathrm{CaCO}_{3}\) increase, decrease, or remain the same as the system approaches equilibrium? (a) \(15.0 \mathrm{~g} \mathrm{CaCO}_{3}, 15.0 \mathrm{~g} \mathrm{CaO},\) and \(4.25 \mathrm{~g} \mathrm{CO}_{2}\) (b) \(2.50 \mathrm{~g} \mathrm{CaCO}_{3}, 25.0 \mathrm{~g} \mathrm{CaO},\) and \(5.66 \mathrm{~g} \mathrm{CO}_{2}\) (a) \(30.5 \mathrm{~g} \mathrm{CaCO}_{3}, 25.5 \mathrm{~g} \mathrm{CaO},\) and \(6.48 \mathrm{~g} \mathrm{CO}_{2}\)

At \(25^{\circ} \mathrm{C}\), the reaction $$\mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)$$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4}\). What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and $\mathrm{CrO}_{4}{ }^{2-}\( in a saturated solution of \)\mathrm{CaCrO}_{4} ?$

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction $4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\( in a fetus, compared to \)K_{c}$ for the same reaction in an adult.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free