Answer :
In order to determine the overall reaction, let's add the given chemical equations and see what cancels out in the process.
The given chemical equations are:
1. \(2 \text{NO}_2(\text{g}) \rightarrow 2 \text{NO}(\text{g}) + \text{O}_2(\text{g})\)
2. \(2 \text{NO}(\text{g}) \rightarrow \text{N}_2(\text{g}) + \text{O}_2(\text{g})\)
3. \(\text{N}_2(\text{g}) + 2 \text{O}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\)
Let's add these equations step-by-step:
First, rewrite equations to add them:
1. \(2 \text{NO}_2(\text{g}) \rightarrow 2 \text{NO}(\text{g}) + \text{O}_2(\text{g})\)
2. \(2 \text{NO}(\text{g}) \rightarrow \text{N}_2(\text{g}) + \text{O}_2(\text{g})\)
3. \(\text{N}_2(\text{g}) + 2 \text{O}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\)
Now, sum the reactants and products of all three equations:
Reactants:
[tex]\[ 2 \text{NO}_2(\text{g}) + 2 \text{NO}(\text{g}) + \text{N}_2(\text{g}) + 2 \text{O}_2(\text{g}) \][/tex]
Products:
[tex]\[ 2 \text{NO}(\text{g}) + \text{O}_2(\text{g}) + \text{N}_2(\text{g}) + \text{O}_2(\text{g}) + \text{N}_2\text{O}_4(\text{g}) \][/tex]
Now, let’s look at what cancels out:
- The \(2 \text{NO}(\text{g})\) on both sides
- The \(\text{N}_2(\text{g})\) on both sides
- The \(\text{O}_2(\text{g})\) from the one in first and second reaction will contribute to cancelling out one \(\text{O}_2(\text{g})\) in the products.
This leaves us with:
- \(2 \text{NO}_2(\text{g})\)
- Production of \(\text{N}_2 \text{O}_4(\text{g})\)
- Remaining \(2 \text{O}_2(\text{g})\) cancels two \(\text{O}_2(\text{g})\) in product side
So, the overall reaction is:
[tex]\[2 \text{NO}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\][/tex]
This matches with the balanced overall reaction:
[tex]\[2 \text{NO}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\][/tex]
The given chemical equations are:
1. \(2 \text{NO}_2(\text{g}) \rightarrow 2 \text{NO}(\text{g}) + \text{O}_2(\text{g})\)
2. \(2 \text{NO}(\text{g}) \rightarrow \text{N}_2(\text{g}) + \text{O}_2(\text{g})\)
3. \(\text{N}_2(\text{g}) + 2 \text{O}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\)
Let's add these equations step-by-step:
First, rewrite equations to add them:
1. \(2 \text{NO}_2(\text{g}) \rightarrow 2 \text{NO}(\text{g}) + \text{O}_2(\text{g})\)
2. \(2 \text{NO}(\text{g}) \rightarrow \text{N}_2(\text{g}) + \text{O}_2(\text{g})\)
3. \(\text{N}_2(\text{g}) + 2 \text{O}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\)
Now, sum the reactants and products of all three equations:
Reactants:
[tex]\[ 2 \text{NO}_2(\text{g}) + 2 \text{NO}(\text{g}) + \text{N}_2(\text{g}) + 2 \text{O}_2(\text{g}) \][/tex]
Products:
[tex]\[ 2 \text{NO}(\text{g}) + \text{O}_2(\text{g}) + \text{N}_2(\text{g}) + \text{O}_2(\text{g}) + \text{N}_2\text{O}_4(\text{g}) \][/tex]
Now, let’s look at what cancels out:
- The \(2 \text{NO}(\text{g})\) on both sides
- The \(\text{N}_2(\text{g})\) on both sides
- The \(\text{O}_2(\text{g})\) from the one in first and second reaction will contribute to cancelling out one \(\text{O}_2(\text{g})\) in the products.
This leaves us with:
- \(2 \text{NO}_2(\text{g})\)
- Production of \(\text{N}_2 \text{O}_4(\text{g})\)
- Remaining \(2 \text{O}_2(\text{g})\) cancels two \(\text{O}_2(\text{g})\) in product side
So, the overall reaction is:
[tex]\[2 \text{NO}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\][/tex]
This matches with the balanced overall reaction:
[tex]\[2 \text{NO}_2(\text{g}) \rightarrow \text{N}_2\text{O}_4(\text{g})\][/tex]