Question

which of the following represents a formation equation for c₆h₁₂o₆(s)?
○ 6 c(gr) + 12 h₂(g) + 6 o₂(g) ⟶ c₆h₁₂o₆(s)
○ c₆h₁₂o₆(s) ⟶ 6 c(gr) + 6 h₂(g) + 3 o₂(g)
○ 6 c(s) + 6 h₂(g) + 3/2 o₂(g) ⟶ c₆h₁₂o₆(s)
○ 6 c(gr) + 6 h₂(g) + 3 o₂(g) ⟶ c₆h₁₂o₆(s)

Explanation:

Step1: Recall Formation Reaction Definition

A formation reaction is when 1 mole of a compound is formed from its elements in their standard states. Carbon's standard state is graphite (gr) or solid C, \( \ce{H2(g)} \) is hydrogen's standard state, \( \ce{O2(g)} \) is oxygen's standard state. The formula for glucose is \( \ce{C6H12O6} \), so we need 6 C, 12 H (from \( 6\ce{H2} \)), and 6 O (from \( 3\ce{O2} \))? Wait, no: let's balance atoms. For \( \ce{C6H12O6} \), C: 6, H: 12 (so \( 6\ce{H2} \) since each \( \ce{H2} \) has 2 H), O: 6 (so \( 3\ce{O2} \) since each \( \ce{O2} \) has 2 O) or \( 3/2\ce{O2} \)? Wait, no: let's check each option.

Step2: Analyze Each Option

- Option 1: \( 6\ce{C(gr)} + 12\ce{H2(g)} + 6\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). H: 12*2=24, but glucose has 12 H. So H is over. Wrong.
- Option 2: It's a decomposition, not formation (reactant is glucose, product is elements). Formation is elements to compound. Wrong.
- Option 3: \( 6\ce{C(s)} \) (but standard C is graphite, but sometimes written as C(s) or C(gr)). H: 6\( \ce{H2} \) gives 12 H (good). O: \( 3/2\ce{O2} \) gives 3 O? Wait no, \( 3/2\ce{O2} \) has 3 O atoms? Wait \( \ce{O2} \) is diatomic, so \( 3/2\ce{O2} \) has 3 O atoms? No, \( 3/2\ce{O2} \) has \( 3/2 * 2 = 3 \) O atoms? Wait glucose has 6 O. Oh wait, I messed up. Wait glucose is \( \ce{C6H12O6} \), so O: 6. So \( 3\ce{O2} \) would give 6 O (since \( 3\ce{O2} \) has 6 O atoms). Wait no, let's recalculate:

Wait for \( \ce{C6H12O6} \), number of O atoms: 6. So from \( \ce{O2} \), each \( \ce{O2} \) has 2 O, so moles of \( \ce{O2} \) needed: \( 6/2 = 3 \) moles? Wait no, wait the formula:

Let's balance the formation reaction:

\( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). Wait:

C: 6 (good), H: 6*2=12 (good), O: 3*2=6 (good). Wait but option 4: \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). Wait no, option 4 is \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \)? Wait no, the fourth option is \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). Wait but let's check option 4:

Wait the fourth option: \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). Let's count atoms:

C: 6 (good), H: 6*2=12 (good), O: 3*2=6 (good). Wait but option 3 has \( 3/2\ce{O2} \), which would give 3 O atoms (since \( 3/2 * 2 = 3 \)), but glucose has 6 O. So option 3 is wrong. Wait I must have made a mistake. Wait glucose formula: \( \ce{C6H12O6} \), so O: 6. So moles of \( \ce{O2} \) needed: 6 O atoms / 2 O per \( \ce{O2} \) = 3 moles of \( \ce{O2} \). Wait no: 1 mole \( \ce{O2} \) has 2 moles O atoms. So to get 6 moles O atoms, need 3 moles \( \ce{O2} \). Wait but option 4: \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). Let's check H: 6\( \ce{H2} \) gives 12 H (good, glucose has 12 H). C: 6 C (good). O: 3\( \ce{O2} \) gives 6 O (good, glucose has 6 O). Wait but option 3: \( 6\ce{C(s)} + 6\ce{H2(g)} + 3/2\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). O: \( 3/2\ce{O2} \) has 3 O atoms (since \( 3/2 * 2 = 3 \)), but glucose has 6 O. So option 3 is wrong. Wait the fourth option: \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). Wait no, the fourth option in the image is \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \)? Wait no, let's recheck the options:

Wait the options are:

1. \( 6\ce{C(gr)} + 12\ce{H2(g)} + 6\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \) (H: 24, wrong)
2. Decomposition (wrong)
3. \( 6\ce{C(s)} + 6\ce{H2(g)} + 3/2\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \) (O: 3, but glucose needs 6 O? Wait no, glucose is \( \ce{C6H12O6} \), so O: 6. So \( 3/2\ce{O2} \) has 3 O atoms? No, \( 3/2\ce{O2} \) is 1.5 moles of \( \ce{O2} \), which has 3 moles of O atoms. But glucose has 6 moles of O atoms. So this is wrong. Wait I'm confused. Wait maybe the third option has a typo, or I misread. Wait no, let's calculate the number of O atoms in glucose: \( \ce{C6H12O6} \) has 6 O atoms. So from \( \ce{O2} \), each \( \ce{O2} \) molecule has 2 O atoms. So moles of \( \ce{O2} \) needed: 6 O atoms / 2 O per \( \ce{O2} \) = 3 moles of \( \ce{O2} \). Wait but the third option has \( 3/2\ce{O2} \), which is 1.5 moles, giving 3 O atoms. That's half of what's needed. So that's wrong. Then the fourth option: \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). Let's check H: 6\( \ce{H2} \) gives 12 H (good). C: 6 C (good). O: 3\( \ce{O2} \) gives 6 O (good). Wait but the third option: maybe the C is graphite (gr) but written as C(s). Wait maybe the third option is wrong, and the fourth is correct? Wait no, let's check the original options again. Wait the third option is \( 6\ce{C(s)} + 6\ce{H2(g)} + 3/2\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). Wait H: 6*2=12 (good). C: 6 (good). O: \( 3/2\ce{O2} \) has 3 O atoms? No, \( 3/2\ce{O2} \) is 1.5 moles of \( \ce{O2} \), which has 3 moles of O atoms. But glucose has 6 moles of O atoms. So that's wrong. Then the fourth option: \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). H: 6*2=12 (good). C: 6 (good). O: 3*2=6 (good). Wait but the first option: \( 12\ce{H2} \) gives 24 H, which is wrong. So the correct option should be the fourth one? Wait no, wait the formula for glucose: \( \ce{C6H12O6} \), so H is 12, so \( 6\ce{H2} \) (each \( \ce{H2} \) has 2 H) gives 12 H (correct). O: 6 O, so \( 3\ce{O2} \) (each \( \ce{O2} \) has 2 O) gives 6 O (correct). C: 6 C (correct). So the fourth option: \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). Wait but let's check the third option again. Wait maybe I made a mistake in O count. Wait \( \ce{C6H12O6} \): O is 6, so moles of O: 6. From \( \ce{O2} \), moles of \( \ce{O2} \) needed: 6/2 = 3 moles. So \( 3\ce{O2} \) is correct. So the fourth option is correct? Wait no, the fourth option in the image is \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \)? Wait no, looking back:

Wait the options are:

1. \( 6\ce{C(gr)} + 12\ce{H2(g)} + 6\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \) (H: 24, wrong)
2. \( \ce{C6H12O6(s)} ightarrow 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} \) (decomposition, wrong)
3. \( 6\ce{C(s)} + 6\ce{H2(g)} + 3/2\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \) (O: 3, wrong)
4. \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \) (H: 12, C:6, O:6, correct)

Wait but wait, the third option: maybe the C is graphite, and the O is \( 3/2\ce{O2} \), but that gives 3 O, but glucose has 6 O. So no. So the fourth option is correct. Wait but let's check the number of H atoms. \( 6\ce{H2} \) has 12 H atoms (6*2), which matches glucose's 12 H. C: 6 C, matches. O: 3\( \ce{O2} \) has 6 O atoms (3*2), matches glucose's 6 O. So the fourth option is correct. Wait but in the options, the fourth is \( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)} ightarrow \ce{C6H12O6(s)} \). So that's the correct formation equation.

Answer:

\( 6\ce{C(gr)} + 6\ce{H2(g)} + 3\ce{O2(g)}
ightarrow \ce{C6H12O6(s)} \) (the fourth option)