Chapter 12, Problem 64QAP

### Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373

Chapter
Section

### Chemistry: Principles and Reactions

8th Edition
William L. Masterton + 1 other
ISBN: 9781305079373
Textbook Problem

# For the decomposition of CaCO3 at 900°C, K = 1.04 . CaCO 3 ( s ) ⇌ CaO ( s ) + O 2 ( g ) Find the smallest mass of CaCO3 needed to reach equilibrium in a 5.00-L vessel at 900°C.

Interpretation Introduction

Interpretation:

The mass of calcium carbonate required to reach equilibrium needs to be calculated.

Concept introduction:

For a general equilibrium reaction as follows:

A(g)+B(g)C(g)+D(g)

The expression for the equilibrium constant is represented as follows:

K=(PC)(PD)(PA)(PB)

Here, to calculate the equilibrium constant, the values of partial pressure of all the species in reactant and product side are required.

The number of moles of gas can be calculated from pressure as follows:

n=PVRT

Here, P is pressure, V is volume, R is Universal gas constant and T is temperature.

The mass of a substance can be calculated from number of moles and mass as follows:

m=n×M

Here, n is number of moles and M is molar mass.

Explanation

The given equilibrium reaction is as follows:

CaCO3(s)â‡ŒCaO(s)+O2(g)

The value of equilibrium constant is given 1.04.

From the equilibrium equation, the expression for equilibrium constant will be:

K=PO2

Therefore,

PO2=1.04Â atm

The number of moles of oxygen gas can be calculated as follows:

n=PVRT

The volume of the vessel is 5.00 L at 900Â âˆ˜C

The temperature can be converted into K as follows:

0Â âˆ˜C=273.15Â K

Or,

900Â âˆ˜C=900+273.15Â K=1173.15Â K

Putting the values,

n=(1

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