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Sagot :
To determine the enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction:
[tex]\[ 2 \text{C (graphite)} + \text{H}_2 (\text{g}) \rightarrow \text{C}_2\text{H}_2 (\text{g}), \][/tex]
we need to manipulate the given reactions and their enthalpy changes. Here are the given reactions with their enthalpy changes:
(a) [tex]\( \text{C (graphite)} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \)[/tex] [tex]\(\Delta H^\circ = -393.5 \, \text{kJ}\)[/tex].
(b) [tex]\( \text{H}_2 (\text{g}) + \frac{1}{2} \text{O}_2 (\text{g}) \rightarrow \text{H}_2 \text{O (l)} \)[/tex] [tex]\(\Delta H^\circ = -285.8 \, \text{kJ}\)[/tex].
(c) [tex]\( 2 \text{C}_2\text{H}_2 (\text{g}) + 5 \text{O}_2 (\text{g}) \rightarrow 4 \text{CO}_2 (\text{g}) + 2 \text{H}_2\text{O (l)} \)[/tex] [tex]\(\Delta H^\circ = -2598.8 \, \text{kJ}\)[/tex].
First, we want to find the enthalpy change for the target reaction, [tex]\( 2 \text{C (graphite)} + \text{H}_2 (\text{g}) \rightarrow \text{C}_2\text{H}_2 (\text{g}) \)[/tex]. To do this, we need to somehow manipulate the given reactions to end up with our target reaction.
Let's analyze what we need to do:
1. We start by manipulating reaction (c):
- We need to reverse reaction (c) because in reaction (c), [tex]\(\text{C}_2\text{H}_2\)[/tex] is on the reactant side, but we need it on the product side.
- When reversing a reaction, the sign of [tex]\(\Delta H^\circ\)[/tex] changes.
Reversed reaction (c) becomes:
[tex]\[ 4 \text{CO}_2 (\text{g}) + 2 \text{H}_2 \text{O (l)} \rightarrow 2 \text{C}_2\text{H}_2 (\text{g}) + 5 \text{O}_2 (\text{g}), \][/tex]
and the enthalpy change for this reaction is:
[tex]\[ \Delta H^\circ = +2598.8 \, \text{kJ}. \][/tex]
2. We divide the entire reversed reaction (c) by 2 to match the stoichiometry of our target reaction:
[tex]\[ 2 \text{CO}_2 (\text{g}) + \text{H}_2\text{O (l)} \rightarrow \text{C}_2\text{H}_2 (\text{g}) + \frac{5}{2} \text{O}_2 (\text{g}), \][/tex]
and the enthalpy change for the divided reaction is:
[tex]\[ \Delta H^\circ = \frac{+2598.8 \, \text{kJ}}{2} = +1299.4 \, \text{kJ}. \][/tex]
3. Summing up the manipulated equations:
Our manipulated equation now directly matches the stoichiometry of our desired target reaction:
[tex]\[ 2 \text{C (graphite)} + \text{H}_2 (\text{g}) \rightarrow \text{C}_2\text{H}_2 (\text{g}), \][/tex]
with the enthalpy change:
[tex]\[ \Delta H^\circ = +1299.4 \, \text{kJ}. \][/tex]
Therefore, the enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction [tex]\(2 \text{C (graphite)} + \text{H}_2 (\text{g}) \rightarrow \text{C}_2\text{H}_2 (\text{g})\)[/tex] is:
[tex]\[ \Delta H^\circ = +1299.4 \, \text{kJ}. \][/tex]
[tex]\[ 2 \text{C (graphite)} + \text{H}_2 (\text{g}) \rightarrow \text{C}_2\text{H}_2 (\text{g}), \][/tex]
we need to manipulate the given reactions and their enthalpy changes. Here are the given reactions with their enthalpy changes:
(a) [tex]\( \text{C (graphite)} + \text{O}_2 (\text{g}) \rightarrow \text{CO}_2 (\text{g}) \)[/tex] [tex]\(\Delta H^\circ = -393.5 \, \text{kJ}\)[/tex].
(b) [tex]\( \text{H}_2 (\text{g}) + \frac{1}{2} \text{O}_2 (\text{g}) \rightarrow \text{H}_2 \text{O (l)} \)[/tex] [tex]\(\Delta H^\circ = -285.8 \, \text{kJ}\)[/tex].
(c) [tex]\( 2 \text{C}_2\text{H}_2 (\text{g}) + 5 \text{O}_2 (\text{g}) \rightarrow 4 \text{CO}_2 (\text{g}) + 2 \text{H}_2\text{O (l)} \)[/tex] [tex]\(\Delta H^\circ = -2598.8 \, \text{kJ}\)[/tex].
First, we want to find the enthalpy change for the target reaction, [tex]\( 2 \text{C (graphite)} + \text{H}_2 (\text{g}) \rightarrow \text{C}_2\text{H}_2 (\text{g}) \)[/tex]. To do this, we need to somehow manipulate the given reactions to end up with our target reaction.
Let's analyze what we need to do:
1. We start by manipulating reaction (c):
- We need to reverse reaction (c) because in reaction (c), [tex]\(\text{C}_2\text{H}_2\)[/tex] is on the reactant side, but we need it on the product side.
- When reversing a reaction, the sign of [tex]\(\Delta H^\circ\)[/tex] changes.
Reversed reaction (c) becomes:
[tex]\[ 4 \text{CO}_2 (\text{g}) + 2 \text{H}_2 \text{O (l)} \rightarrow 2 \text{C}_2\text{H}_2 (\text{g}) + 5 \text{O}_2 (\text{g}), \][/tex]
and the enthalpy change for this reaction is:
[tex]\[ \Delta H^\circ = +2598.8 \, \text{kJ}. \][/tex]
2. We divide the entire reversed reaction (c) by 2 to match the stoichiometry of our target reaction:
[tex]\[ 2 \text{CO}_2 (\text{g}) + \text{H}_2\text{O (l)} \rightarrow \text{C}_2\text{H}_2 (\text{g}) + \frac{5}{2} \text{O}_2 (\text{g}), \][/tex]
and the enthalpy change for the divided reaction is:
[tex]\[ \Delta H^\circ = \frac{+2598.8 \, \text{kJ}}{2} = +1299.4 \, \text{kJ}. \][/tex]
3. Summing up the manipulated equations:
Our manipulated equation now directly matches the stoichiometry of our desired target reaction:
[tex]\[ 2 \text{C (graphite)} + \text{H}_2 (\text{g}) \rightarrow \text{C}_2\text{H}_2 (\text{g}), \][/tex]
with the enthalpy change:
[tex]\[ \Delta H^\circ = +1299.4 \, \text{kJ}. \][/tex]
Therefore, the enthalpy change ([tex]\(\Delta H\)[/tex]) for the reaction [tex]\(2 \text{C (graphite)} + \text{H}_2 (\text{g}) \rightarrow \text{C}_2\text{H}_2 (\text{g})\)[/tex] is:
[tex]\[ \Delta H^\circ = +1299.4 \, \text{kJ}. \][/tex]
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