Q97CP

Question

Bromine monochloride is formed from the elements:

$∆ H_{r x n}^{°} = - {1.35 kJ/mol ∆G}_{r x n}^{°} =-0.88 kJ/mol$

Calculate

(a) $∆ H_{f}^{°}$ and

(b) $S^{°}$ of $BrCl(g)$

Step-by-Step Solution

Verified

Answer

The value of $∆ H_{f}^{°}$ for $BrCl(g)$ is $14.78 k J / m o l$ .
The value of $S^{°}$ for $BrCl(g)$ is $237.9 k J / m o l \cdot K$ .

1Step 1: Concept Introduction

Gibbs free energy, also known as Gibbs function, Gibbs energy, or free enthalpy, is a term used to measure the greatest amount of work done in a thermodynamic system when temperature and pressure remain constant.

2Step 2: Calculation for ∆ H f ° ∆ H f °

(a)

The balanced equation is –

${Cl}_{2} {(g) + Br}_{2} (g) \to 2BrCl(g)$

The values given are –

$∆ H_{r x n}^{°} = - 1.35 kJ/mol ∆ G_{f}^{°} = - 0.88 kJ/mol$

Manipulate the reaction to solve for $∆ H_{f}^{°}$ of $BrCl(g)$ –

$∆ H_{r x n}^{°} = \sum_{}^{} n_{p} ∆ H_{f}^{°} (p r o d u c t) - \sum_{}^{} n_{r} ∆ H_{f}^{°} (r e a c \tan t) ∆ H_{r x n}^{°} = [n ∆ H_{f}^{°} (BrCl (g))] - [n ∆ H_{f}^{°} (C l_{2} (g)) + n ∆ H_{f}^{°} (B r_{2 (g)})] n ∆ H_{f}^{°} (B r_{2} (g)) = [n ∆ H_{f}^{°} (C l_{2} (g)) + n ∆ H_{f}^{°} (B r_{2} (g))] + ∆ H_{r x n}^{°} 2 ∆ H_{f}^{°} (B r C l (g)) = [1 (0) + 1 (30.91)] k J / m o l + (1.35 k J / m o l) ∆ H_{f}^{°} (B r C l (g)) = \frac{29.56 k J / m o l}{2} ∆ H_{f}^{°} (B r C l (g)) = 14.78 k J / m o l$

Therefore, the value is obtained as $4.78 k J / m o l$ .

3Step 3: Calculation for S °

(b)

Solve to obtain the value for $∆ G_{r x n}^{°}$ first –

$∆ G_{r x n}^{°} = \sum_{}^{} n_{p} ∆ G_{f}^{°} (p r o d u c t) - \sum_{}^{} n_{r} ∆ G_{f}^{°} (r e a c \tan t) = [n ∆ G_{f}^{°} (B r C l (g))] - [n ∆ G_{f}^{°} (C l_{2} (g)) + n ∆ G_{f}^{°} (B r_{2} (g))] = ([2 (- 0.88)] - [1 (0) + 1 (3.13)]) k J / m o l ∆ G_{r x n}^{°} = - 4.89 k J / m o l$

Now, solve for $∆ S_{r x n}^{°}$ at $298 K$ –

$∆ G_{r x n}^{°} = ∆ H_{r x n}^{°} - T ∆ S_{r x n}^{°} T ∆ S_{r x n}^{°} = ∆ H_{r x n}^{°} - ∆ G_{r x n}^{°} ∆ S_{r x n}^{°} = \frac{∆ H_{_{r x n}}^{°} - ∆ G_{r x n}^{°}}{T} = \frac{(2 \times (- 1.35 k J / m o l)) - (- 4.89 k J / m o l)}{298 K} ∆ S_{r x n}^{°} = 0.00735 k J / m o l = 7.35 J / m o l \cdot K$

Now manipulate the equation to solve for $S^{°}$ of $B r C l (g)$ –

$∆ S^{°} = {\sum n_{p}}_{}^{} S^{°} (p r o d u c t) - \sum_{}^{} n_{r} S^{°} (r e a c \tan t) ∆ S^{°} = [n S^{°} (B r C l (g))] - [n S^{°} (B r_{2} (g)) + n S^{°} (C l_{2} (g))] n S^{°} (B r C l (g)) = ∆ S^{°} + [n S^{°} (B r_{2} (g)) + n S^{°} (C l_{2} (g))] 2 S^{°} (B r C l (g)) = (7.35 . J / m o l \cdot K) + ([1 (223.0) + 1 (245.38)]) J / m o l \cdot K S^{°} (B r C l (g)) = \frac{475.73 J / m o l \cdot K}{2} S^{°} (B r C l (g)) = 237.9 j / m o l \cdot K$

Therefore, the value is obtained as $237.9 k J / m o l \cdot K$ .

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