Answer :
Sure, let's evaluate the given equations step by step to determine the correct one for calculating [tex]\(\Delta H_{\text{rxn}}\)[/tex].
Equation 1:
[tex]\[ (-393.5 \, \text{kJ/mol} + (-285.83 \, \text{kJ/mol})) - (-1273.02 \, \text{kJ/mol}) \][/tex]
Summing the terms inside the first parenthesis:
[tex]\[ -393.5 + (-285.83) = -393.5 - 285.83 = -679.33 \, \text{kJ/mol} \][/tex]
Then subtracting the term outside:
[tex]\[ -679.33 - (-1273.02) = -679.33 + 1273.02 = 593.69 \, \text{kJ/mol} \][/tex]
So, the result for Equation 1 is:
[tex]\[ 593.69 \, \text{kJ/mol} \][/tex]
Equation 2:
[tex]\[ -1273.02 \, \text{kJ/mol} - ((6 \, \text{mol}) \times (-393.5 \, \text{kJ/mol}) + (6 \, \text{mol}) \times (-285.83 \, \text{kJ/mol})) \][/tex]
First, multiply the values inside the parentheses:
[tex]\[ 6 \times (-393.5) = -2361 \, \text{kJ/mol} \][/tex]
[tex]\[ 6 \times (-285.83) = -1714.98 \, \text{kJ/mol} \][/tex]
Sum these two results:
[tex]\[ -2361 + (-1714.98) = -2361 - 1714.98 = -4075.98 \, \text{kJ/mol} \][/tex]
Then subtract this from [tex]\(-1273.02 \, \text{kJ/mol}\)[/tex]:
[tex]\[ -1273.02 - (-4075.98) = -1273.02 + 4075.98 = 2802.96 \, \text{kJ/mol} \][/tex]
So, the result for Equation 2 is:
[tex]\[ 2802.96 \, \text{kJ/mol} \][/tex]
Equation 3:
[tex]\[ ((6 \, \text{mol}) \times (-393.5 \, \text{kJ/mol}) + (6 \, \text{mol}) \times (-285.83 \, \text{kJ/mol})) - (1 \, \text{mol}) \times (-1273.02 \, \text{kJ/mol}) \][/tex]
The terms inside the parentheses have already been calculated:
[tex]\[ -4075.98 \, \text{kJ/mol} \][/tex]
Then subtract the last term:
[tex]\[ -4075.98 - (-1273.02) = -4075.98 + 1273.02 = -2802.96 \, \text{kJ/mol} \][/tex]
So, the result for Equation 3 is:
[tex]\[ -2802.96 \, \text{kJ/mol} \][/tex]
Considering the results:
- Equation 1: [tex]\(593.69 \, \text{kJ/mol}\)[/tex]
- Equation 2: [tex]\(2802.96 \, \text{kJ/mol}\)[/tex]
- Equation 3: [tex]\(-2802.96 \, \text{kJ/mol}\)[/tex]
Therefore, based on the results and the calculation of [tex]\(\Delta H_{\text{rxn}}\)[/tex], the equation that correctly reflects the enthalpy change for this reaction is:
[tex]\[ (-393.5 \, \text{kJ/mol} + (-285.83 \, \text{kJ/mol})) - (-1273.02 \, \text{kJ/mol}) \][/tex]
This matches the result [tex]\(593.69 \, \text{kJ/mol}\)[/tex] from Equation 1.
Equation 1:
[tex]\[ (-393.5 \, \text{kJ/mol} + (-285.83 \, \text{kJ/mol})) - (-1273.02 \, \text{kJ/mol}) \][/tex]
Summing the terms inside the first parenthesis:
[tex]\[ -393.5 + (-285.83) = -393.5 - 285.83 = -679.33 \, \text{kJ/mol} \][/tex]
Then subtracting the term outside:
[tex]\[ -679.33 - (-1273.02) = -679.33 + 1273.02 = 593.69 \, \text{kJ/mol} \][/tex]
So, the result for Equation 1 is:
[tex]\[ 593.69 \, \text{kJ/mol} \][/tex]
Equation 2:
[tex]\[ -1273.02 \, \text{kJ/mol} - ((6 \, \text{mol}) \times (-393.5 \, \text{kJ/mol}) + (6 \, \text{mol}) \times (-285.83 \, \text{kJ/mol})) \][/tex]
First, multiply the values inside the parentheses:
[tex]\[ 6 \times (-393.5) = -2361 \, \text{kJ/mol} \][/tex]
[tex]\[ 6 \times (-285.83) = -1714.98 \, \text{kJ/mol} \][/tex]
Sum these two results:
[tex]\[ -2361 + (-1714.98) = -2361 - 1714.98 = -4075.98 \, \text{kJ/mol} \][/tex]
Then subtract this from [tex]\(-1273.02 \, \text{kJ/mol}\)[/tex]:
[tex]\[ -1273.02 - (-4075.98) = -1273.02 + 4075.98 = 2802.96 \, \text{kJ/mol} \][/tex]
So, the result for Equation 2 is:
[tex]\[ 2802.96 \, \text{kJ/mol} \][/tex]
Equation 3:
[tex]\[ ((6 \, \text{mol}) \times (-393.5 \, \text{kJ/mol}) + (6 \, \text{mol}) \times (-285.83 \, \text{kJ/mol})) - (1 \, \text{mol}) \times (-1273.02 \, \text{kJ/mol}) \][/tex]
The terms inside the parentheses have already been calculated:
[tex]\[ -4075.98 \, \text{kJ/mol} \][/tex]
Then subtract the last term:
[tex]\[ -4075.98 - (-1273.02) = -4075.98 + 1273.02 = -2802.96 \, \text{kJ/mol} \][/tex]
So, the result for Equation 3 is:
[tex]\[ -2802.96 \, \text{kJ/mol} \][/tex]
Considering the results:
- Equation 1: [tex]\(593.69 \, \text{kJ/mol}\)[/tex]
- Equation 2: [tex]\(2802.96 \, \text{kJ/mol}\)[/tex]
- Equation 3: [tex]\(-2802.96 \, \text{kJ/mol}\)[/tex]
Therefore, based on the results and the calculation of [tex]\(\Delta H_{\text{rxn}}\)[/tex], the equation that correctly reflects the enthalpy change for this reaction is:
[tex]\[ (-393.5 \, \text{kJ/mol} + (-285.83 \, \text{kJ/mol})) - (-1273.02 \, \text{kJ/mol}) \][/tex]
This matches the result [tex]\(593.69 \, \text{kJ/mol}\)[/tex] from Equation 1.
