Question

which of the following best represents a formation equation?
c(graphite) + 3/2 h₂(g) + 1/2 cl₂(g) → ch₃cl(g)
c(s) + 3/2 h₂(g) + cl₂ → ch₃cl(g)
c + 3/2 h₂ + 1/2 cl₂ → ch₃cl
2 c(graphite) + 3 h₂(g) + cl₂(g) → 2 ch₃cl(g)

Explanation:

A formation equation for a compound shows the formation of 1 mole of the compound from its constituent elements in their standard states. Carbon's standard state is graphite (solid), hydrogen's is \( \text{H}_2(\text{g}) \), and chlorine's is \( \text{Cl}_2(\text{g}) \). We need 1 mole of \( \text{CH}_3\text{Cl}(\text{g}) \) formed.

• Option 1: \( \text{C}(\text{graphite}) + \frac{3}{2}\text{H}_2(\text{g}) + \frac{1}{2}\text{Cl}_2(\text{g})

ightarrow \text{CH}_3\text{Cl}(\text{g}) \) uses elements in standard states to form 1 mol of \( \text{CH}_3\text{Cl} \).

• Option 2: Uses \( \text{C}(\text{s}) \) (same as graphite, but notation; however, the product is 1 mol, but let's check others).
• Option 3: Uses \( \text{C} \) (not standard state notation; standard C is graphite/solid, but written as C, ambiguous, and standard Cl is \( \text{Cl}_2 \), not \( \text{Cl} \)).
• Option 4: Forms 2 moles of \( \text{CH}_3\text{Cl} \), but formation should be 1 mole.

So Option 1 is correct as it forms 1 mol of \( \text{CH}_3\text{Cl} \) from elements in standard states.

Answer:

\( \boldsymbol{\text{C}(\text{graphite}) + \frac{3}{2}\text{H}_2(\text{g}) + \frac{1}{2}\text{Cl}_2(\text{g})
ightarrow \text{CH}_3\text{Cl}(\text{g})} \) (the first option in the list)