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Sagot :
Let's calculate the standard enthalpy change for each reaction step-by-step. We will use known standard enthalpies of formation (ΔH_f°) for each substance involved.
### Reaction (a):
[tex]\[ H_2(g) + \frac{1}{2} O_2(g) \longrightarrow H_2O(l) \][/tex]
To find the enthalpy change (ΔH) for this reaction, we use the standard enthalpies of formation for the reactants and products. The standard enthalpy of formation (ΔH_f°) is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states.
Given:
- ΔH_f° (H_2O(l)) = -285.8 kJ/mol
- ΔH_f° (H_2(g)) = 0 kJ/mol (standard enthalpy of formation for any element in its standard state is zero)
- ΔH_f° (O_2(g)) = 0 kJ/mol (same reasoning as above)
The enthalpy change for the reaction can be calculated using:
[tex]\[ \Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) \][/tex]
For reaction (a):
[tex]\[ \Delta H^\circ = [\Delta H_f^\circ(\text{H}_2\text{O(l)})] - [\Delta H_f^\circ(\text{H}_2(g)) + \frac{1}{2} \Delta H_f^\circ(\text{O}_2(g))] \][/tex]
[tex]\[ \Delta H^\circ = [-285.8 \, \text{kJ/mol}] - [0 + 0] \][/tex]
[tex]\[ \Delta H^\circ = -285.8 \, \text{kJ/mol} \][/tex]
### Reaction (b):
[tex]\[ C_2H_4(g) + 3O_2(g) \longrightarrow C_2H_5OH(l) \][/tex]
For this reaction, we need the standard enthalpies of formation for ethylene (C_2H_4), ethanol (C_2H_5OH), and the standard states of carbon dioxide (CO_2) and water (H_2O).
Given:
- ΔH_f° (C_2H_5OH(l)) = -277.0 kJ/mol
- ΔH_f° (C_2H_4(g)) = 52.26 kJ/mol
- ΔH_f° (O_2(g)) = 0 kJ/mol (standard state)
- ΔH_f° (CO_2(g)) = -393.5 kJ/mol
- ΔH_f° (H_2O(l)) = -285.8 kJ/mol
Again, using the enthalpy change formula:
[tex]\[ \Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) \][/tex]
For reaction (b):
[tex]\[ \Delta H^\circ = [\Delta H_f^\circ(\text{C}_2\text{H}_5\text{OH(l)})] - [\Delta H_f^\circ(\text{C}_2\text{H}_4(g)) + 3 \Delta H_f^\circ(\text{O}_2(g))] \][/tex]
[tex]\[ \Delta H^\circ = [-277.0 \, \text{kJ/mol}] - [52.26 \, \text{kJ/mol} + 3 \times 0 \, \text{kJ/mol}] \][/tex]
[tex]\[ \Delta H^\circ = -277.0 \, \text{kJ/mol} - 52.26 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H^\circ = -329.26 \, \text{kJ/mol} \][/tex]
### Summary of Results:
1. The standard enthalpy change for reaction (a):
[tex]\[ \Delta H = -285.8 \, \text{kJ/mol} \][/tex]
2. The standard enthalpy change for reaction (b):
[tex]\[ \Delta H = -329.26 \, \text{kJ/mol} \][/tex]
### Reaction (a):
[tex]\[ H_2(g) + \frac{1}{2} O_2(g) \longrightarrow H_2O(l) \][/tex]
To find the enthalpy change (ΔH) for this reaction, we use the standard enthalpies of formation for the reactants and products. The standard enthalpy of formation (ΔH_f°) is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states.
Given:
- ΔH_f° (H_2O(l)) = -285.8 kJ/mol
- ΔH_f° (H_2(g)) = 0 kJ/mol (standard enthalpy of formation for any element in its standard state is zero)
- ΔH_f° (O_2(g)) = 0 kJ/mol (same reasoning as above)
The enthalpy change for the reaction can be calculated using:
[tex]\[ \Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) \][/tex]
For reaction (a):
[tex]\[ \Delta H^\circ = [\Delta H_f^\circ(\text{H}_2\text{O(l)})] - [\Delta H_f^\circ(\text{H}_2(g)) + \frac{1}{2} \Delta H_f^\circ(\text{O}_2(g))] \][/tex]
[tex]\[ \Delta H^\circ = [-285.8 \, \text{kJ/mol}] - [0 + 0] \][/tex]
[tex]\[ \Delta H^\circ = -285.8 \, \text{kJ/mol} \][/tex]
### Reaction (b):
[tex]\[ C_2H_4(g) + 3O_2(g) \longrightarrow C_2H_5OH(l) \][/tex]
For this reaction, we need the standard enthalpies of formation for ethylene (C_2H_4), ethanol (C_2H_5OH), and the standard states of carbon dioxide (CO_2) and water (H_2O).
Given:
- ΔH_f° (C_2H_5OH(l)) = -277.0 kJ/mol
- ΔH_f° (C_2H_4(g)) = 52.26 kJ/mol
- ΔH_f° (O_2(g)) = 0 kJ/mol (standard state)
- ΔH_f° (CO_2(g)) = -393.5 kJ/mol
- ΔH_f° (H_2O(l)) = -285.8 kJ/mol
Again, using the enthalpy change formula:
[tex]\[ \Delta H = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants}) \][/tex]
For reaction (b):
[tex]\[ \Delta H^\circ = [\Delta H_f^\circ(\text{C}_2\text{H}_5\text{OH(l)})] - [\Delta H_f^\circ(\text{C}_2\text{H}_4(g)) + 3 \Delta H_f^\circ(\text{O}_2(g))] \][/tex]
[tex]\[ \Delta H^\circ = [-277.0 \, \text{kJ/mol}] - [52.26 \, \text{kJ/mol} + 3 \times 0 \, \text{kJ/mol}] \][/tex]
[tex]\[ \Delta H^\circ = -277.0 \, \text{kJ/mol} - 52.26 \, \text{kJ/mol} \][/tex]
[tex]\[ \Delta H^\circ = -329.26 \, \text{kJ/mol} \][/tex]
### Summary of Results:
1. The standard enthalpy change for reaction (a):
[tex]\[ \Delta H = -285.8 \, \text{kJ/mol} \][/tex]
2. The standard enthalpy change for reaction (b):
[tex]\[ \Delta H = -329.26 \, \text{kJ/mol} \][/tex]
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