
Calculate the partial pressure of carbon monoxide from the following:
$CaCO_{3(s)}\overset{\Delta}{\rightarrow}CaO_{(s)}+CO_{2(g)}\uparrow$ ${{K}_{p}}=8\times {{10}^{-2}}$
$CO_{2(s)}+C_{(s)}\rightarrow 2CO_{(s)}$ ${{K}_{p}}=2$
(A) 0.2
(B) 0.4
(C) 1.6
(D) 4
Answer
164.1k+ views
Hint: When a chemical reaction reaches equilibrium, the equilibrium constant gives information about the interaction between the reactants and products. ${{K}_{p}}$ is the equilibrium constant produced by partial pressures in a reaction equation. It is used to describe the interaction between reactant and product pressures. Temperature and the stoichiometry of the change both affect ${{K}_{p}}$ values. ${{K}_{p}}$ is measured in ${{(atm)}^{\Delta n}}$ units.
Complete step by step solution:
The first equation given is
$CaCO_{3(s)}\overset{\Delta}{\rightarrow}CaO_{(s)}+CO_{2(g)}\uparrow$
As in this reaction, only $C{{O}_{2}}$is a gas. So, the ${{K}_{p}}$given in the question is the partial pressure of$C{{O}_{2}}$. ${{P}_{C{{O}_{2}}}}=8\times {{10}^{-2}}$ …(1)
Now, the second reaction given is:
$CO_{2(s)}+C_{(s)}\rightarrow 2CO_{(s)}$
The equilibrium constant is calculated as ${{K}_{p}}=\frac{{{\left( {{P}_{CO}} \right)}^{2}}}{{{P}_{C{{O}_{2}}}}}$
As ${{K}_{p}}$for this reaction given in the question is 2. So, $2=\frac{{{\left( {{P}_{CO}} \right)}^{2}}}{{{P}_{C{{O}_{2}}}}}$ … (2)
From equations (1) and (2),
$2=\frac{{{({{P}_{CO}})}^{2}}}{8\times {{10}^{-2}}}$
${{({{P}_{CO}})}^{2}}=2\times 8\times {{10}^{-2}}$
${{({{P}_{CO}})}^{2}}=16\times {{10}^{-2}}$
${{({{P}_{CO}})}^{2}}=0.16$
${{P}_{CO}}=0.4$
Hence, the partial pressure of carbon monoxide ($CO$) is 0.4.
Correct Option: (B) 0.4.
Note: Partial pressure is the pressure that one gas in a gas mixture will exert if it occupies the same volume on its own. In a mixture, every gas exerts a particular pressure. The thermodynamic activity of a gas is determined by its partial pressure. The overall pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases in the mixture, according to Dalton's equation of partial pressure. For an ideal gas mixture, Dalton's law is completely true.
Complete step by step solution:
The first equation given is
$CaCO_{3(s)}\overset{\Delta}{\rightarrow}CaO_{(s)}+CO_{2(g)}\uparrow$
As in this reaction, only $C{{O}_{2}}$is a gas. So, the ${{K}_{p}}$given in the question is the partial pressure of$C{{O}_{2}}$. ${{P}_{C{{O}_{2}}}}=8\times {{10}^{-2}}$ …(1)
Now, the second reaction given is:
$CO_{2(s)}+C_{(s)}\rightarrow 2CO_{(s)}$
The equilibrium constant is calculated as ${{K}_{p}}=\frac{{{\left( {{P}_{CO}} \right)}^{2}}}{{{P}_{C{{O}_{2}}}}}$
As ${{K}_{p}}$for this reaction given in the question is 2. So, $2=\frac{{{\left( {{P}_{CO}} \right)}^{2}}}{{{P}_{C{{O}_{2}}}}}$ … (2)
From equations (1) and (2),
$2=\frac{{{({{P}_{CO}})}^{2}}}{8\times {{10}^{-2}}}$
${{({{P}_{CO}})}^{2}}=2\times 8\times {{10}^{-2}}$
${{({{P}_{CO}})}^{2}}=16\times {{10}^{-2}}$
${{({{P}_{CO}})}^{2}}=0.16$
${{P}_{CO}}=0.4$
Hence, the partial pressure of carbon monoxide ($CO$) is 0.4.
Correct Option: (B) 0.4.
Note: Partial pressure is the pressure that one gas in a gas mixture will exert if it occupies the same volume on its own. In a mixture, every gas exerts a particular pressure. The thermodynamic activity of a gas is determined by its partial pressure. The overall pressure of a mixture of gases is equal to the sum of the partial pressures of the constituent gases in the mixture, according to Dalton's equation of partial pressure. For an ideal gas mixture, Dalton's law is completely true.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

Displacement-Time Graph and Velocity-Time Graph for JEE

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

Instantaneous Velocity - Formula based Examples for JEE

Thermodynamics Class 11 Notes: CBSE Chapter 5
