Chapter 2: Q103CP (page 40)

From the following reaction and data, find (a) $S^{°} o f S O C l_{2}$ and (b) T at which the reaction becomes nonspontaneous:
${SO}_{3} {(g) + SCl}_{2} (l \to {SOCl}_{2} {(l) + SO}_{2} {(g)∆G}_{r x n}^{°} =-75.2kJ {SO}_{3} {(g)SCl}_{2} {(l)SOCl}_{2} {(l)SO}_{2} (g) ∆ H_{f}^{°} (k J / m o l) - 396 - 50.0 - 245.6 - 296.8 S^{°} (J / m o l \times K) 256.7 184 - 248.1$

Short Answer

Expert verified

a) The entropy of $S O C l_{2}$ production is $192.529 J / m o l \times K$ .

b) Because $∆ G_{r x n}^{°} > 0$ , the reaction is non-spontaneous at temperatures above $1.355 \times 10^{6} K$ .

Step by step solution

Definition of Nonspontaneous reaction

A chemical reaction known as an endergonic reaction (also known as a nonspontaneous reaction) is one in which the standard change in free energy is positive and energy is absorbed. By linking endergonic processes to highly exergonic reactions, they can be pushed or pulled.

CalculatingS° of SOCl2

a) Because entropy is a state function, it can be stated in the following way:

$∆ S_{r x n}^{°} = \sum_{}^{} ∆ S_{p r o d u c t}^{°} - \sum_{}^{} ∆ S_{r e a c \tan t}^{°}$

In the case of the stated reaction,

$∆ S_{r x n}^{°} = [1 m o l \times ∆ S_{S O C l_{2} (I)}^{°} + 1 m o l \times ∆ S_{S O_{2} (g)}^{°}] - - [1 m o l \times ∆ S_{S O_{3} (g)}^{°} + 1 m o l \times ∆ S_{S C l_{2} (I)}^{°}]$

As a result, the entropy of the $S O C l_{2}$ can be written as

$∆ S_{r x n}^{°} = [1 m o l \times ∆ S_{S O C l_{2}}^{°} + 1 m o l \times ∆ S_{S O_{2 (g)}}^{°}] - - [1 m o l \times ∆ S_{S O_{3} (g)}^{°} + 1 m o l \times ∆ S_{S C l_{2} (I)}^{°}] ∆ S_{S O C l_{2} (I)}^{°} = ∆ S_{r x n}^{°} + [1 m o l \times ∆ S_{S O_{3} (g)}^{°} + 1 m o l \times ∆ S_{S C l_{2} (I)}^{°}] - [1 m o l \times ∆ S_{S O_{2} (g)}^{°}]$

The reaction's total entropy is not given, but it can be calculated:

$∆ G_{r x n}^{°} = ∆ H_{r x n}^{°} - T \times ∆ S_{r x n}^{°} ∆ S_{r x n}^{°} = \frac{∆ H_{r x n}^{°} - ∆ G_{r x n}^{°}}{T}$

Calculating reaction’s total enthalpy

The reaction's total enthalpy can be calculated as follows:

$∆ H_{r x n}^{°} = [1 m o l \times ∆ H_{S O C l_{2}}^{°} + 1 m o l \times ∆ H_{S O_{2} (g)}^{°}] - - [1 m o l \times ∆ H_{S O_{3} (g)}^{°} + 1 m o l \times ∆ H_{S C l_{2} (I)}^{°}] ∆ H_{r x n}^{°} = [1 m o l \times (- 245.60 k J / m o l)_1 m o l \times (- 296.80 k J / m o l)] - - [1 m o l \times (- 396.00 k J / m o l) + 1 m o l \times (- 50.00 k J / m o l)] ∆ H_{r x n}^{°} = - 96.40 k J$
Now we can calculate the total reaction entropy:

$∆ S_{r x n}^{°} = \frac{- 96.40 k J - (- 75.20 k J)}{298 K} ∆ S_{r x n}^{°} = - 0.07114 J / K = - 7.114 \times 10^{- 2} J / K$

As a result, the standard entropy of a reaction is $- 7.114 \times 10^{- 2} J / K$ .

Finally, the entropy of $S O C l_{2}$ can be determined as follows:

$∆ S_{S O C l_{2} (I)}^{°} = - 7.114 \times 10^{- 2} \frac{J}{m o l \times K} + + [1 m o l \times 256.70 \frac{J}{m o l \times K} + 1 m o l \times 184.00 \frac{J}{m o l \times K}] - - [1 m o l \times 248.10 \frac{J}{m o l \times K}] ∆ S_{S O C l_{2} (I)}^{°} = 192.529 J / m o l \times K$

$S O C l_{2}$ has a formation entropy of $192.259 J / m o l \times K$ .

Calculating T at which reaction become nonspontaneous

b) We must apply the following equation to determine the temperature at which the reaction becomes non-spontaneous: $∆ G_{r x n}^{°} = ∆ H_{r x n}^{°} - T \times ∆ S_{r x n}^{°}$

When $∆ G_{r x n}^{°} > 0$ , the reaction becomes non-spontaneous at T. To begin, determine the temperature at which $∆ G_{r x n}^{°} = 0$ exists.

$∆ G_{r x n}^{°} = ∆ H_{r x n}^{°} - T \times ∆ S_{r x n}^{°} ∆ H_{r x n}^{°} - T \times ∆ S_{r x n}^{°} = 0 T = \frac{∆ H_{r x n}^{°}}{∆ S_{r x n}^{°}}$

Using the values determined in a) for normal reaction circumstances

$T = \frac{- 96.40 \times 10^{3} J}{- 7.114 \times 10^{- 2} J / K} T = 1.355 \times 10^{6} K$

Because the reaction is at equilibrium at $1.355 \times 10^{6} K$ , the forward and backward reactions have the same rate. The reaction is non-spontaneous above that temperature because $∆ G_{r x n}^{°} > 0$ .

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	Mass (g) of Sample	Mass(g) of Sulfur in Sample
Coal A	378	11.3
Coal B	495	19.0
Coal C	675	20.6

Short Answer

Step by step solution

Definition of Nonspontaneous reaction

CalculatingS° of SOCl2

Calculating reaction’s total enthalpy

Calculating T at which reaction become nonspontaneous

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Making Measurements

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From the following reaction and data, find (a)S°ofSOCl2and (b) T at which the reaction becomes nonspontaneous:SO3(g) + SCl2(l→SOCl2(l) + SO2(g)∆Grxn°=-75.2kJSO3(g)SCl2(l)SOCl2(l)SO2(g)∆Hf°(kJ/mol)-396-50.0-245.6-296.8S°(J/mol×K)256.7184-248.1

Short Answer

Step by step solution

Definition of Nonspontaneous reaction

CalculatingS° of SOCl2

Calculating reaction’s total enthalpy

Calculating T at which reaction become nonspontaneous

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Chemistry Textbooks

Making Measurements

Chemical Reactions

Chemical Analysis

Kinetics

Chemistry Branches

Ionic and Molecular Compounds

Study anywhere. Anytime. Across all devices.