Chapter 12, Problem 62QAP

### 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

# Mustard gas, used in chemical warfare in World War I, has been found to be an effective agent in the chemotherapy of Hodgkin's disease. It can be produced according to the following reaction: SCl 2 ( g ) + 2 C 2 H 4 ( g ) ⇌ S ( CH 2 CH 2 Cl ) 2 ( g ) An evacuated 5.0-1- flask at 20.0°C is filled with 0.258 mol SCl2 and 0.592 mol C2H4. After equilibrium is established, 0.0349 mol mustard gas is present.(a) What is the partial pressure of each gas at equilibrium?(b) What is K at 20.0°C?

Interpretation Introduction

(a)

Interpretation:

The equilibrium partial pressure of each gas needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

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.

Explanation

The volume of the flask is 5.0 L at 20âˆ˜C.

The temperature can be converted into K as follows:

0Â âˆ˜C=273.15Â K

Or,

20Â âˆ˜C=20+273.15Â K=293.15Â K

The initial number of moles of SCl2 and C2H4 is 0.258 mol and 0.592 mol respectively.

At equilibrium, the number of moles of mustard gas or S(CH2CH2Cl)2(g) is 0.0349 mol.

The initial pressure of SCl2 can be calculated as follows:

P=nRTV

Putting the values,

PSCl2,initial=(0.258Â mol)(0.082Â LÂ atm/KÂ mol)(293.15Â K)(5.0Â L)=1.24Â atm

Similarly, initial pressure of C2H4 can be calculated as follows:

PC2H4,initial=(0.592Â mol)(0.082Â LÂ atm/KÂ mol)(293.15Â K)(5.0Â L)=2.846Â atm

Now, partial pressure of S(CH2CH2Cl)2(g) at equilibrium will be:

PS(CH2CH2Cl)2=(0.0349Â mol)(0.082Â LÂ atm/KÂ mol)(293.15Â K)(5

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant for the given temperature needs to be determined.

Concept introduction:

The system is said to be in equilibrium if the there is no change in the partial pressure or concentration of reactant and product takes place.

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.

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