Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
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.
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
