To find the overall reaction from the given reaction mechanism, we will sum up all the steps and then cancel out any intermediates that appear on both the reactant and product sides.
Given reaction mechanism:
1. [tex]\( 2A \rightarrow B + 2C \)[/tex] (slow)
2. [tex]\( B + C \rightarrow D + E \)[/tex] (fast)
3. [tex]\( C + D \rightarrow E + F \)[/tex] (fast)
Step-by-step process:
1. Sum up all the reactants and products:
- Reactants: [tex]\( 2A \)[/tex], [tex]\( B \)[/tex], [tex]\( C \)[/tex], [tex]\( C \)[/tex], [tex]\( D \)[/tex]
- Products: [tex]\( B \)[/tex], [tex]\( 2C \)[/tex], [tex]\( D \)[/tex], [tex]\( E \)[/tex], [tex]\( E \)[/tex], [tex]\( F \)[/tex]
This gives us:
- Total reactants: [tex]\( 2A + B + 2C + C + D \)[/tex]
- Total products: [tex]\( B + 2C + D + E + E + F \)[/tex]
2. Combine like terms:
- Reactants: [tex]\( 2A + B + 3C + D \)[/tex]
- Products: [tex]\( B + 2C + D + 2E + F \)[/tex]
3. Cancel out intermediates:
- Identify intermediates that appear on both sides: [tex]\( B \)[/tex], [tex]\( 2C \)[/tex], [tex]\( D \)[/tex]
- Cancel these terms out:
- Reactants remaining: [tex]\( 2A \)[/tex]
- Products remaining: [tex]\( 2E + F \)[/tex]
Therefore, the overall reaction is:
[tex]\[ 2A \rightarrow 2E + F \][/tex]
So the final, simplified overall reaction is:
[tex]\[ \boxed{2A \rightarrow 2E + F} \][/tex]
