Question
Use the information in the table to answer the question.%0D%0A%0D%0ABond%09Bond Energy (kJ/mol)%0D%0AH-I%09297%0D%0AH-H%09435%0D%0A I-I%09151%0D%0AHydrogen and iodine gases react to produce hydroiodic acid according to the reaction.%0D%0A%0D%0AH2+I2→2HI%0D%0A%0D%0AHow much total bond energy in kJ/mol is in the products of the reaction?%0D%0A%0D%0A(1 point)%0D%0AResponses%0D%0A%0D%0A453%0D%0A453%0D%0A%0D%0A297%0D%0A297%0D%0A%0D%0A594%0D%0A594%0D%0A%0D%0A448%0D%0A448
Answers
GPT-4o mini
To determine the total bond energy in the products of the reaction \( \text{H}_2 + \text{I}_2 \rightarrow 2\text{HI} \), we need to calculate the bond energies for the bonds formed in the products (2 HI).
The bond energy for H-I is given as 297 kJ/mol. Since there are 2 HI molecules produced in the reaction, the total bond energy for the products is calculated as follows:
\[
\text{Total bond energy in products} = 2 \times \text{Bond energy of H-I}
\]
\[
= 2 \times 297 \, \text{kJ/mol}
\]
\[
= 594 \, \text{kJ/mol}
\]
Therefore, the total bond energy in the products of the reaction is **594 kJ/mol**.
The bond energy for H-I is given as 297 kJ/mol. Since there are 2 HI molecules produced in the reaction, the total bond energy for the products is calculated as follows:
\[
\text{Total bond energy in products} = 2 \times \text{Bond energy of H-I}
\]
\[
= 2 \times 297 \, \text{kJ/mol}
\]
\[
= 594 \, \text{kJ/mol}
\]
Therefore, the total bond energy in the products of the reaction is **594 kJ/mol**.