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 (1/3)c₆h₁₂o₆(s) → ch₄(g) + co₂(g) kj 3ch₄(g) + 3co₂(g) → c₆h₁₂o₆(s) kj 6ch₄(g) + 6co₂(g) → 2c₆h₁₂o₆(s) kj

Explanation:

Step1: Analyze the first reaction

The original reaction is \( \ce{C6H12O6(s) -> 3CH4(g) + 3CO2(g)} \) with \( \Delta H = -132\ \text{kJ} \). For the reaction \( \frac{1}{3}\ce{C6H12O6(s) -> CH4(g) + CO2(g)} \), we divide the original reaction by 3. So we also divide \( \Delta H \) by 3.
\( \Delta H_1=\frac{-132}{3}=-44\ \text{kJ} \)

Step2: Analyze the second reaction

The second reaction is the reverse of the original reaction: \( 3\ce{CH4(g)} + 3\ce{CO2(g)} -> \ce{C6H12O6(s)} \). When we reverse a reaction, the sign of \( \Delta H \) changes. So \( \Delta H_2 = -(-132)=132\ \text{kJ} \)

Step3: Analyze the third reaction

The third reaction is \( 6\ce{CH4(g)} + 6\ce{CO2(g)} -> 2\ce{C6H12O6(s)} \). This is twice the reverse of the original reaction (multiply the reversed original reaction by 2). So we multiply the \( \Delta H \) of the reversed reaction by 2. The reversed original reaction has \( \Delta H = 132\ \text{kJ} \), so \( \Delta H_3 = 2\times132 = 264\ \text{kJ} \)? Wait, no, wait. Wait the original reaction is \( \ce{C6H12O6(s) -> 3CH4(g) + 3CO2(g)} \) \( \Delta H=-132 \). The reaction \( 6\ce{CH4(g)} + 6\ce{CO2(g)} -> 2\ce{C6H12O6(s)} \) is 2 times the reverse of the original reaction. The reverse of original is \( 3\ce{CH4} + 3\ce{CO2} -> \ce{C6H12O6} \) with \( \Delta H = 132 \). Then multiplying by 2: \( 6\ce{CH4} + 6\ce{CO2} -> 2\ce{C6H12O6} \), so \( \Delta H = 2\times132 = 264 \)? Wait no, wait the original \( \Delta H \) is -132. Reverse reaction \( \Delta H = +132 \). Then multiplying by 2: \( \Delta H = 2\times132 = 264 \)? Wait, no, wait the third reaction is \( 6\ce{CH4} + 6\ce{CO2} -> 2\ce{C6H12O6} \). The original reaction is 1 mole of glucose produces 3 moles of \( \ce{CH4} \) and 3 moles of \( \ce{CO2} \). The third reaction is 6 moles of \( \ce{CH4} \) and 6 moles of \( \ce{CO2} \) producing 2 moles of glucose. So that's 2 times the reverse of the original reaction (since original is 1 glucose to 3 \( \ce{CH4} \) and 3 \( \ce{CO2} \); reverse is 3 \( \ce{CH4} \) and 3 \( \ce{CO2} \) to 1 glucose; multiply by 2: 6 \( \ce{CH4} \) and 6 \( \ce{CO2} \) to 2 glucose). So the \( \Delta H \) for the reversed original is +132, multiply by 2: \( \Delta H = 2\times132 = 264 \)? Wait, no, wait the original \( \Delta H \) is -132. So reverse is +132. Then multiplying by 2: \( \Delta H = 2\times132 = 264 \). Wait but let's check again.

Wait the first reaction: original is \( \ce{C6H12O6} -> 3\ce{CH4} + 3\ce{CO2} \) \( \Delta H=-132 \). The reaction \( \frac{1}{3}\ce{C6H12O6} -> \ce{CH4} + \ce{CO2} \): divide original by 3, so \( \Delta H = -132/3 = -44 \). Correct.

Second reaction: reverse of original: \( 3\ce{CH4} + 3\ce{CO2} -> \ce{C6H12O6} \), so \( \Delta H = +132 \). Correct.

Third reaction: \( 6\ce{CH4} + 6\ce{CO2} -> 2\ce{C6H12O6} \). This is 2 times the second reaction (since second reaction is 3 \( \ce{CH4} \) and 3 \( \ce{CO2} \) to 1 glucose; multiply by 2: 6 \( \ce{CH4} \) and 6 \( \ce{CO2} \) to 2 glucose). So \( \Delta H = 2\times132 = 264 \)? Wait, no, wait the second reaction's \( \Delta H \) is 132, so multiplying by 2 gives 264. Wait but let's check the stoichiometry. Original reaction: 1 glucose -> 3 CH4 + 3 CO2. The third reaction: 6 CH4 + 6 CO2 -> 2 glucose. So that's 2*(3 CH4 + 3 CO2 -> 1 glucose) which is 2 times the reversed original reaction. So \( \Delta H = 2*132 = 264 \). Wait but the original \( \Delta H \) is -132, so reversed is +132, times 2 is +264. Wait but let's confirm.

Wait the first reaction: \( \frac{1}{3}\ce{C6H12O6} -> \ce{CH4} + \ce{CO2} \): \( \Delta H = -132/3 = -44 \). Correct.

Second reaction: reverse of original: \( 3\ce{CH4} + 3\ce{CO2} -> \ce{C6H12O6} \), \( \Delta H = +132 \). Correct.

Third reaction: \( 6\ce{CH4} + 6\ce{CO2} -> 2\ce{C6H12O6} \). This is 2*(3\ce{CH4} + 3\ce{CO2} -> \ce{C6H12O6}), so \( \Delta H = 2*132 = 264 \). Wait, but wait, the original reaction's \( \Delta H \) is -132, so when we reverse it, we get +132, then multiply by 2, we get +264. So that's correct.

Wait but let's check the third reaction again. The original reaction is 1 glucose to 3 CH4 and 3 CO2. The third reaction is 6 CH4 and 6 CO2 to 2 glucose. So the moles of CH4 and CO2 are doubled, and moles of glucose are doubled. So it's 2 times the reverse of the original reaction. So yes, \( \Delta H = 2*132 = 264 \).

Wait but wait, the user's table has three reactions. Let's list the answers:

First reaction: -44 kJ

Second reaction: +132 kJ

Third reaction: 2*132 = 264 kJ? Wait no, wait the original reaction is \( \ce{C6H12O6} -> 3\ce{CH4} + 3\ce{CO2} \) \( \Delta H=-132 \). The reaction \( 6\ce{CH4} + 6\ce{CO2} -> 2\ce{C6H12O6} \) is the reverse of 2 times the original reaction. Wait, original reaction times 2: \( 2\ce{C6H12O6} -> 6\ce{CH4} + 6\ce{CO2} \) \( \Delta H = 2*(-132) = -264 \). Then reverse that reaction: \( 6\ce{CH4} + 6\ce{CO2} -> 2\ce{C6H12O6} \) \( \Delta H = +264 \). Ah, that's another way to see it. So original times 2: \( 2\ce{C6H12O6} -> 6\ce{CH4} + 6\ce{CO2} \) \( \Delta H = -264 \). Then reverse: \( 6\ce{CH4} + 6\ce{CO2} -> 2\ce{C6H12O6} \) \( \Delta H = +264 \). Yes, that's correct. So the third reaction's \( \Delta H \) is 264 kJ.

So summarizing:

1. \( \frac{1}{3}\ce{C6H12O6} -> \ce{CH4} + \ce{CO2} \): \( \Delta H = -132/3 = -44 \)

2. \( 3\ce{CH4} + 3\ce{CO2} -> \ce{C6H12O6} \): \( \Delta H = +132 \)

3. \( 6\ce{CH4} + 6\ce{CO2} -> 2\ce{C6H12O6} \): \( \Delta H = +264 \)

Wait, but let's confirm the third reaction again. Original reaction: 1 glucose -> 3 CH4 + 3 CO2, \( \Delta H=-132 \). If we want 6 CH4 + 6 CO2 -> 2 glucose, that's the reverse of 2 glucose -> 6 CH4 + 6 CO2. The reaction 2 glucose -> 6 CH4 + 6 CO2 is 2 times the original reaction, so \( \Delta H = 2*(-132) = -264 \). Then reverse that: 6 CH4 + 6 CO2 -> 2 glucose, \( \Delta H = +264 \). Yes, that's correct. So the third reaction's \( \Delta H \) is 264 kJ.

Answer:

First reaction \( \Delta H \): \(-44\) kJ
Second reaction \( \Delta H \): \(132\) kJ
Third reaction \( \Delta H \): \(264\) kJ

(If the table requires filling each box: first box -44, second box 132, third box 264)