Question

a chemist measures the enthalpy change δh during the following reaction: c₆h₁₂o₆(s)→2co₂(g) + 2c₂h₅oh(l) δh=−69. kj use this information to complete the table below. round each of your answers to the nearest kj. reaction δh 8co₂(g) + 8c₂h₅oh(l) → 4c₆h₁₂o₆(s) kj ½c₆h₁₂o₆(s) → co₂(g) + c₂h₅oh(l) kj 2co₂(g) + 2c₂h₅oh(l) → c₆h₁₂o₆(s) kj

Explanation:

Step1: Analyze the first reaction

The original reaction is \( \ce{C6H12O6(s) -> 2CO2(g) + 2C2H5OH(l)} \) with \( \Delta H = -69\ \text{kJ} \). The first reaction in the table is \( \ce{8CO2(g) + 8C2H5OH(l) -> 4C6H12O6(s)} \). This is the reverse of the original reaction multiplied by 4. So we reverse the sign of \( \Delta H \) and multiply by 4.
\( \Delta H_1 = -(-69\ \text{kJ}) \times 4 = 69 \times 4 = 276\ \text{kJ} \)

Step2: Analyze the second reaction

The second reaction is \( \ce{\frac{1}{2}C6H12O6(s) -> CO2(g) + C2H5OH(l)} \). This is the original reaction multiplied by \( \frac{1}{2} \). So we multiply \( \Delta H \) by \( \frac{1}{2} \).
\( \Delta H_2 = -69\ \text{kJ} \times \frac{1}{2} = -34.5\ \text{kJ} \approx -35\ \text{kJ} \) (rounded to nearest kJ)

Step3: Analyze the third reaction

The third reaction is \( \ce{2CO2(g) + 2C2H5OH(l) -> C6H12O6(s)} \). This is the reverse of the original reaction. So we reverse the sign of \( \Delta H \).
\( \Delta H_3 = -(-69\ \text{kJ}) = 69\ \text{kJ} \)

Answer:

For the first reaction: \( 276\ \text{kJ} \)
For the second reaction: \( -35\ \text{kJ} \)
For the third reaction: \( 69\ \text{kJ} \)