Question

a chemist measures the enthalpy change δh during the following reaction: c₆h₁₂o₆(s)→3ch₄(g) + 3co₂(g) δh = -132. kj use this information to complete the table below. round each of your answers to the nearest kj. reaction δh 3ch₄(g) + 3co₂(g) → c₆h₁₂o₆(s) kj (1/2)ch₄(g) + (1/2)co₂(g) → (1/6)c₆h₁₂o₆(s) kj 3c₆h₁₂o₆(s) → 9ch₄(g) + 9co₂(g) kj

Explanation:

Step1: Reverse the reaction

The original reaction is \( \ce{C6H12O6(s) -> 3CH4(g) + 3CO2(g)} \) with \( \Delta H = -132\space kJ \). For the reaction \( \ce{3CH4(g) + 3CO2(g) -> C6H12O6(s)} \), we reverse the original reaction. When a reaction is reversed, the sign of \( \Delta H \) changes. So \( \Delta H = -(-132\space kJ) = 132\space kJ \).

Step2: Scale the reaction

For the reaction \( \ce{\frac{1}{2}CH4(g) + \frac{1}{2}CO2(g) -> \frac{1}{6}C6H12O6(s)} \), we take the reversed reaction from Step1 (\( \ce{3CH4(g) + 3CO2(g) -> C6H12O6(s)} \), \( \Delta H = 132\space kJ \)) and divide all coefficients by 6. When we divide the coefficients by 6, we also divide \( \Delta H \) by 6. So \( \Delta H = \frac{132\space kJ}{6} = 22\space kJ \) (wait, no, wait: original reversed reaction has \( \Delta H = 132\space kJ \) for 3 moles of \( \ce{CH4} \) and 3 moles of \( \ce{CO2} \) producing 1 mole of \( \ce{C6H12O6} \). The new reaction has \( \frac{1}{2} \) moles of \( \ce{CH4} \) (which is \( \frac{3}{6} \)) and \( \frac{1}{2} \) moles of \( \ce{CO2} \) ( \( \frac{3}{6} \)) producing \( \frac{1}{6} \) moles of \( \ce{C6H12O6} \). So we divide the reversed reaction by 6: \( \frac{1}{6}(3\ce{CH4} + 3\ce{CO2} -> \ce{C6H12O6}) \) gives \( \frac{1}{2}\ce{CH4} + \frac{1}{2}\ce{CO2} -> \frac{1}{6}\ce{C6H12O6} \), and \( \Delta H = \frac{132}{6} = 22\space kJ \)? Wait, no, original reaction \( \ce{C6H12O6 -> 3CH4 + 3CO2} \), \( \Delta H = -132 \). So the reverse is \( \ce{3CH4 + 3CO2 -> C6H12O6} \), \( \Delta H = +132 \). Now, the reaction \( \ce{\frac{1}{2}CH4 + \frac{1}{2}CO2 -> \frac{1}{6}C6H12O6} \): let's see the stoichiometry. The reverse reaction has 3 \( \ce{CH4} \), 3 \( \ce{CO2} \) to 1 \( \ce{C6H12O6} \). The new reaction has \( \frac{1}{2} \) \( \ce{CH4} \) (which is \( 3\times\frac{1}{6} \)), \( \frac{1}{2} \) \( \ce{CO2} \) ( \( 3\times\frac{1}{6} \)) and \( \frac{1}{6} \) \( \ce{C6H12O6} \) ( \( 1\times\frac{1}{6} \)). So we multiply the reverse reaction by \( \frac{1}{6} \). So \( \Delta H = 132\times\frac{1}{6} = 22\space kJ \)? Wait, no, wait: the original reaction is exothermic ( \( \Delta H negative \) ), so the reverse is endothermic ( \( \Delta H positive \) ). But when we take a fraction of the reverse reaction, we scale \( \Delta H \) by that fraction. Wait, maybe I made a mistake earlier. Let's re - express:

Original reaction: \( n = 1\) mole of \( \ce{C6H12O6} \) produces 3 moles \( \ce{CH4} \) and 3 moles \( \ce{CO2} \), \( \Delta H = - 132\space kJ \) (exothermic, releases 132 kJ).

Reverse reaction: 3 moles \( \ce{CH4} \) and 3 moles \( \ce{CO2} \) form 1 mole \( \ce{C6H12O6} \), \( \Delta H = + 132\space kJ \) (endothermic, absorbs 132 kJ).

Now, the reaction \( \ce{\frac{1}{2}CH4 + \frac{1}{2}CO2 -> \frac{1}{6}C6H12O6} \): the amount of reactants is \( \frac{1}{2} \) moles of \( \ce{CH4} \) and \( \ce{CO2} \), which is \( \frac{3}{6} \) moles (since 3 moles in reverse reaction). The product is \( \frac{1}{6} \) moles of \( \ce{C6H12O6} \). So we take the reverse reaction and divide by 6. So \( \Delta H \) for the reverse reaction is 132 kJ for 3 moles reactants and 1 mole product. Dividing by 6: \( \frac{132\space kJ}{6}=22\space kJ \). Wait, but let's check the sign. The reverse reaction is endothermic (positive \( \Delta H \)), and when we take a fraction of it, the \( \Delta H \) for the fraction is also positive? Wait, no, wait: the original reaction is \( \ce{C6H12O6 -> 3CH4 + 3CO2} \), \( \Delta H=-132 \). So the reaction \( \ce{\frac{1}{6}C6H12O6 -> \frac{1}{2}CH4 + \frac{1}{2}CO2} \) would have \( \Delta H=\frac{- 132}{6}=-22\space kJ \). Then reversing this reaction (to get \( \ce{\frac{1}{2}CH4 + \frac{1}{2}CO2 -> \frac{1}{6}C6H12O6} \)) would give \( \Delta H = 22\space kJ \). Yes, that's correct.

Step3: Scale the original reaction

For the reaction \( \ce{3C6H12O6(s) -> 9CH4(g) + 9CO2(g)} \), we take the original reaction \( \ce{C6H12O6(s) -> 3CH4(g) + 3CO2(g)} \) with \( \Delta H=-132\space kJ \) and multiply all coefficients by 3. When we multiply the coefficients by 3, we also multiply \( \Delta H \) by 3. So \( \Delta H = 3\times(-132\space kJ)=-396\space kJ \).

Answer:

For the first reaction \( \ce{3CH4(g) + 3CO2(g) -> C6H12O6(s)} \), \( \Delta H = 132\space kJ \)

For the second reaction \( \ce{\frac{1}{2}CH4(g) + \frac{1}{2}CO2(g) -> \frac{1}{6}C6H12O6(s)} \), \( \Delta H = 22\space kJ \) (Wait, no, wait: Wait, original reaction \( \ce{C6H12O6 -> 3CH4 + 3CO2} \), \( \Delta H=-132 \). The reaction \( \ce{\frac{1}{6}C6H12O6 -> \frac{1}{2}CH4 + \frac{1}{2}CO2} \) has \( \Delta H=\frac{-132}{6}=-22 \). Then reversing it gives \( \ce{\frac{1}{2}CH4 + \frac{1}{2}CO2 -> \frac{1}{6}C6H12O6} \) with \( \Delta H = 22 \). But let's check again. Alternatively, using the reverse reaction from step 1: \( \ce{3CH4 + 3CO2 -> C6H12O6} \), \( \Delta H = 132 \). Dividing by 6: \( \ce{\frac{3}{6}CH4 + \frac{3}{6}CO2 -> \frac{1}{6}C6H12O6} \) which is \( \ce{\frac{1}{2}CH4 + \frac{1}{2}CO2 -> \frac{1}{6}C6H12O6} \), and \( \Delta H=\frac{132}{6}=22 \). Correct.

For the third reaction \( \ce{3C6H12O6(s) -> 9CH4(g) + 9CO2(g)} \), \( \Delta H = 3\times(-132)=-396\space kJ \)

So the answers are 132, 22, - 396 (in the order of the three reactions in the table)