   Chapter 17, Problem 3ALQ ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
1 views

# For the reaction H 2 + I 2 ⇌ 2 HI , consider two possibilities: (a) you add 0.5 mole of each reactant, allow the system to come to equilibrium, and then add 1 mole of H2, and allow the system to reach equilibrium again, or (b) you add 1.5 moles of H2and 0.5 mole of 12 and allow the system to come to equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

Interpretation Introduction

Interpretation:

The final equilibrium mixture for the two procedures is to be explained.

Concept Introduction:

In a chemical equilibrium the concentration of the reactants and products remain constant. The rate of the forward reaction is equal to the rate of backward reaction.

Explanation

The given equation is H2+I22HI.

In the situation a, 0.5 mol of each reactant is added to the system. Let the concentration consumed by each reactant be x. The concentration of the reactants is shown below.

H2  +   I2   2HIInitial conc.(mol)                  0.5       0.5        0Change in conc.(mol)           -x       -x       +2xConc. at equilibrium(mol)    0.5-x   0.5-x  +2x

Then again 1 mole of H2 is added to the system. The equilibrium of the system is disturbed. Therefore, the equilibrium shifts to the right side or in the forward direction. The new equilibrium is established.

Now, it is assumed that the concentration consumed by each reactant be y. Hence, the concentration of each reactant at new equilibrium is shown below.

H2     +    I2        2HI        Conc. at equilibrium(mol)           0.5-x       0.5-x        +2x Conc

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