   Chapter 17, Problem 108CP

Chapter
Section
Textbook Problem

# Consider the reaction H 2 ( g ) + Br 2 ( g ) ⇌ 2HBr ( g ) where ∆H° = −103.8 kJ/mol. In a particular experiment, equal moles of H2(g) at 1.00 atm and Br2(g) at 1.00 atm were mixed in a 1.00-L flask at 25°C and allowed to reach equilibrium. Then the molecules of H2 at equilibrium were counted using a very sensitive technique, and 1.10 × 1013 molecules were found. For this reaction, calculate the values of K, ∆G°, and ∆S°.

Interpretation Introduction

Interpretation: The value of ΔHο for a reaction is given. Equal moles of H2(g) at 1.00 atm and Br2(g) at 1.00 atm were mixed in 1.00 L at 25 οC were mixed and allowed to reach equilibrium. The molecules of H2(g) at equilibrium were counted using a very sensitive technique, and 1.10×1013 molecules were found. The value of K , ΔGο and ΔSο for the reaction is to be calculated.

Concept introduction: Thermodynamics is associated with heat, temperature and its relation with energy and work. It helps us to predict whether a process will take place or not. The terms associated with thermodynamics are system, surrounding, entropy, spontaneity and many more.

Explanation

Explanation

Given

The volume of flask is 1.00 L .

The temperature is 25 οC=298 K .

The pressure is 1 atm .

The ΔHο value is -103.8 kJ/mol .

Equal number of moles of H2(g) and Br2(g) were mixed at a pressure of 1 atm .

The number of moles of H2(g) and Br2(g) is calculated by the formula,

n=PVRT (1)

Where,

• R is the universal gas constant ( 0.08206 Latm/molK ).
• T is the temperature.
• V is the volume.
• P is the pressure.
• n is the number of moles of H2(g) and Br2(g) .

Substitute the value of R , T , V and P in the equation (1).

n=1 atm×1.00 L0.08206 Latm/molK×298 K=0.041 mol

The given reaction is,

H2(g)+Br2(g)2HBr(g)

The change in concentration of H2(g) and Br2(g) is assumed to be x . The ICE table for the given reaction is,

H2(g)+Br2(g)2HBr(g)Initial  0.0410.0410Change xx2xEquilibrium 0.041x0.041x2x

The number of moles of H2(g) present in 6.022×1023 molecules is 1 .

Therefore, the number of moles of H2(g) present in 1×1013 molecules is,

1×1×10136.022×1023=1.66×10-11 Moles of H2(g)

From the ICE table, the number of moles of H2(g) is 0.041x .

Therefore, 0.041x moles of H2(g)=1.66×10-11 Moles of H2(g) .

Similarly, 0.041x moles of Br2(g)=1.66×10-11 Moles of Br2(g) .

Equate 0

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