Chapter 6, Problem 6.12P

(a)

Interpretation Introduction

Interpretation:

The given reaction proceeds to right or left to attain equilibrium has to be predicted.

Concept introduction:

Equilibrium constant (K):

In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product.

Reaction quotient (Q): It is defined as the ratio between concentration of the product and concentration of the reactant.

$\text{Q =}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

This value used to know about the direction of the reaction is going to proceed.

If $\text{Q = K}$ means the reaction reached the equilibrium.

If $\text{Q > K}$ means, the reaction must go to left.

If $\text{Q < K}$ means, the reaction must go to right.

Answer to Problem 6.12P

The reaction should be go to the right.

Explanation of Solution

Given

Given reaction is,

${\text{H}}_{\text{2}\left(\text{g}\right)}{\text{+Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{2HBr}}_{\left(\text{g}\right)}\text{K=7}{\text{.2×10}}^{\text{-4}}$

$\Delta {\text{H}}^o$ is positive.

Amount of $\text{HBr}$ in the vessel = $\text{48}\text{.0}\text{Pa}$

Amount of ${\text{H}}_2$ in the vessel = $1370\text{Pa}$

Amount of ${\text{Br}}_{\text{2}}$ in the vessel = $3310\text{Pa}$

Temperature = $\text{1362}\text{K}$

To calculate: Q value

$\begin{array}{l}\text{Q =}\frac{{\left(\frac{\text{48}\text{.0Pa}}{{\text{10}}^{\text{5}}\text{Pa/bar}}\right)}^{\text{2}}}{\left(\frac{\text{1370Pa}}{{\text{10}}^{\text{5}}\text{Pa/bar}}\right)\left(\frac{\text{3310Pa}}{{\text{10}}^{\text{5}}\text{Pa/bar}}\right)}\\ \\ \text{= 5}{\text{.08×10}}^{\text{-4}}\text{<K}\end{array}$

Where,

$\text{Q < K}$, so the reaction should be go to the right.

(b)

Interpretation Introduction

Interpretation:

The pressure (Pa) of each species at equilibrium has to be calculated.

Answer to Problem 6.12P

$\begin{array}{l}{\text{P}}_{{\text{H}}_{\text{2}}}\text{=1366 Pa}\\ \\ {\text{P}}_{{\text{Br}}_{\text{2}}}\text{=3 306 Pa}\\ \\ {\text{P}}_{\text{HBr}}\text{=57}\text{.0 Pa}\end{array}$

Explanation of Solution

Given

Given reaction is,

${\text{H}}_{\text{2}\left(\text{g}\right)}{\text{+Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{2HBr}}_{\left(\text{g}\right)}\text{K=7}{\text{.2×10}}^{\text{-4}}$

$\Delta {\text{H}}^o$ is positive.

Amount of $\text{HBr}$ in the vessel = $\text{48}\text{.0}\text{Pa}$

Amount of ${\text{H}}_2$ in the vessel = $1370\text{Pa}$

Amount of ${\text{Br}}_{\text{2}}$ in the vessel = $3310\text{Pa}$

Temperature = $\text{1362}\text{K}$

To calculate: pressure (Pa) of each species at equilibrium

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{ + Br}}_{\text{2}}\stackrel{}{\rightleftharpoons }\text{ 2HBr}\\ \\ \text{Initial}\text{pressure:}\text{1370}\text{3310}\text{48}\text{.0}\\ \\ \text{Final}\text{pressure:}\text{1370-x}\text{3310-x}\text{48}\text{.0+2x}\end{array}$

$\begin{array}{l}\frac{{\left(\text{48}\text{.0+2x}\right)}^{\text{2}}}{\left(\text{1370-x}\right)\left(\text{3310-x}\right)}\text{= 7}{\text{.2×10}}^{\text{-4}}\\ \\ \text{x = 4}\text{.50Pa}\end{array}$

So, the value of x has to be subtracted from the final pressure to get pressure.

$\begin{array}{l}{\text{P}}_{{\text{H}}_{\text{2}}}\text{=1366 Pa}\\ \\ {\text{P}}_{{\text{Br}}_{\text{2}}}\text{=3 306 Pa}\\ \\ {\text{P}}_{\text{HBr}}\text{=57}\text{.0 Pa}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The reaction proceed to left or the right to regenerate the equilibrium when the mixture is compressed to half of its original volume has to be explained.

Concept introduction:

Equilibrium constant (K):

In the equilibrium reaction, the ratio of concentration of the reactant and concentration of the product.

Reaction quotient (Q): It is defined as the ratio between concentration of the product and concentration of the reactant.

$\text{Q =}\frac{{\left[\text{C}\right]}^{\text{c}}{\left[\text{D}\right]}^{\text{d}}}{{\left[\text{A}\right]}^{\text{a}}{\left[\text{B}\right]}^{\text{b}}}$

This value used to know about the direction of the reaction is going to proceed.

If $\text{Q = K}$ means the reaction reached the equilibrium.

If $\text{Q > K}$ means, the reaction must go to left.

If $\text{Q < K}$ means, the reaction must go to right.

Explanation of Solution

Given

Given reaction is,

${\text{H}}_{\text{2}\left(\text{g}\right)}{\text{+Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{2HBr}}_{\left(\text{g}\right)}\text{K=7}{\text{.2×10}}^{\text{-4}}$

$\Delta {\text{H}}^o$ is positive.

Amount of $\text{HBr}$ in the vessel = $\text{48}\text{.0}\text{Pa}$

Amount of ${\text{H}}_2$ in the vessel = $1370\text{Pa}$

Amount of ${\text{Br}}_{\text{2}}$ in the vessel = $3310\text{Pa}$

Temperature = $\text{1362}\text{K}$

To explain: reaction proceed to left or the right to regenerate the equilibrium when the mixture is compressed to half of its original volume

The reaction proceeds neither to left nor to right to regenerate the equilibrium when the mixture is compressed to half of its original volume.  Since Q is remains the same.

(d)

Interpretation Introduction

Interpretation:

$\text{HBr}$ be formed or consumed in order to regenerate equilibrium when the mixture at equilibrium is heated from $\text{1362K to 1407K}$ has to be explained.

Concept introduction:

Explanation of Solution

Given

Given reaction is,

${\text{H}}_{\text{2}\left(\text{g}\right)}{\text{+Br}}_{\text{2}\left(\text{g}\right)}\stackrel{}{\rightleftharpoons }{\text{2HBr}}_{\left(\text{g}\right)}\text{K=7}{\text{.2×10}}^{\text{-4}}$

$\Delta {\text{H}}^o$ is positive.

Amount of $\text{HBr}$ in the vessel = $\text{48}\text{.0}\text{Pa}$

Amount of ${\text{H}}_2$ in the vessel = $1370\text{Pa}$

Amount of ${\text{Br}}_{\text{2}}$ in the vessel = $3310\text{Pa}$

Temperature = $\text{1362}\text{K}$

To explain: $\text{HBr}$ be formed or consumed in order to regenerate equilibrium when the mixture at equilibrium is heated from $\text{1362K to 1407K}$

$\text{HBr}$ will be formed in order to regenerate equilibrium when the mixture at equilibrium is heated from $\text{1362K to 1407K}$.  If $\text{HBr}$ is consumed, where the equilibrium cannot regenerate.  Since, $\Delta {\text{H}}^o$ is positive.