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Sagot :
To determine the enthalpy change when 2.50 mol of nitrogen dioxide ([tex]\(NO_2\)[/tex]) decomposes, we need to follow these steps:
1. Understand the Reaction:
The given balanced chemical equation for the decomposition of nitrogen dioxide is:
[tex]\[ 2 \, NO_2(g) \rightarrow N_2(g) + 2 \, O_2(g) \][/tex]
2. Given Data:
- Enthalpy change for the decomposition of [tex]\(NO_2\)[/tex] per mole, [tex]\(\Delta H_t = 33.84 \, \text{kJ/mol}\)[/tex]
- Moles of [tex]\(NO_2\)[/tex] decomposing, [tex]\(\text{moles}_{NO_2} = 2.50 \, \text{mol}\)[/tex]
3. Relate the Entalpy Change to the Reaction:
The enthalpy change given is for the reaction involving 2 moles of [tex]\(NO_2\)[/tex].
4. Calculate the Enthalpy Change:
Since the given enthalpy change ([tex]\(\Delta H_t\)[/tex]) is for the decomposition of 2 moles of [tex]\(NO_2\)[/tex], we can find the enthalpy change for any amount of [tex]\(NO_2\)[/tex] by using a proportion based on that 2 moles.
The enthalpy change for the decomposition of 2 moles of [tex]\(NO_2\)[/tex] is:
[tex]\[ \Delta H_{reaction} = \frac{\text{moles}_{NO_2}}{2} \times \Delta H_t \times (-1) \][/tex]
Here, we multiply by [tex]\(-1\)[/tex] because the reaction is decomposing (which usually releases energy, thus a negative sign).
Plugging in the given values, we get:
[tex]\[ \Delta H_{reaction} = \frac{2.50}{2} \times 33.84 \, \text{kJ/mol} \times (-1) \][/tex]
[tex]\[ \Delta H_{reaction} = 1.25 \times 33.84 \, \text{kJ/mol} \times (-1) \][/tex]
[tex]\[ \Delta H_{reaction} = 42.3 \, \text{kJ} \times (-1) \][/tex]
[tex]\[ \Delta H_{reaction} = -42.3 \, \text{kJ} \][/tex]
5. Interpret the Result:
Since [tex]\(\Delta H_{reaction}\)[/tex] is negative, this indicates that energy is released during the reaction.
Thus, the enthalpy change when 2.50 mol of nitrogen dioxide decomposes is [tex]\(42.3 \, \text{kJ}\)[/tex] of energy released.
Consequently, the correct option is:
[tex]\[ 42.3 \, \text{kJ of energy released} \][/tex]
None of the provided options (13.5 kJ of energy released, 13.5 kJ of energy absorbed, 84.6 kJ of energy released, 84.6 kJ of energy absorbed) match exactly 42.3 kJ of energy released, but given the calculated value, our confidence is that:
\[ \boxed{-42.3 \, \text{kJ of energy released}}
1. Understand the Reaction:
The given balanced chemical equation for the decomposition of nitrogen dioxide is:
[tex]\[ 2 \, NO_2(g) \rightarrow N_2(g) + 2 \, O_2(g) \][/tex]
2. Given Data:
- Enthalpy change for the decomposition of [tex]\(NO_2\)[/tex] per mole, [tex]\(\Delta H_t = 33.84 \, \text{kJ/mol}\)[/tex]
- Moles of [tex]\(NO_2\)[/tex] decomposing, [tex]\(\text{moles}_{NO_2} = 2.50 \, \text{mol}\)[/tex]
3. Relate the Entalpy Change to the Reaction:
The enthalpy change given is for the reaction involving 2 moles of [tex]\(NO_2\)[/tex].
4. Calculate the Enthalpy Change:
Since the given enthalpy change ([tex]\(\Delta H_t\)[/tex]) is for the decomposition of 2 moles of [tex]\(NO_2\)[/tex], we can find the enthalpy change for any amount of [tex]\(NO_2\)[/tex] by using a proportion based on that 2 moles.
The enthalpy change for the decomposition of 2 moles of [tex]\(NO_2\)[/tex] is:
[tex]\[ \Delta H_{reaction} = \frac{\text{moles}_{NO_2}}{2} \times \Delta H_t \times (-1) \][/tex]
Here, we multiply by [tex]\(-1\)[/tex] because the reaction is decomposing (which usually releases energy, thus a negative sign).
Plugging in the given values, we get:
[tex]\[ \Delta H_{reaction} = \frac{2.50}{2} \times 33.84 \, \text{kJ/mol} \times (-1) \][/tex]
[tex]\[ \Delta H_{reaction} = 1.25 \times 33.84 \, \text{kJ/mol} \times (-1) \][/tex]
[tex]\[ \Delta H_{reaction} = 42.3 \, \text{kJ} \times (-1) \][/tex]
[tex]\[ \Delta H_{reaction} = -42.3 \, \text{kJ} \][/tex]
5. Interpret the Result:
Since [tex]\(\Delta H_{reaction}\)[/tex] is negative, this indicates that energy is released during the reaction.
Thus, the enthalpy change when 2.50 mol of nitrogen dioxide decomposes is [tex]\(42.3 \, \text{kJ}\)[/tex] of energy released.
Consequently, the correct option is:
[tex]\[ 42.3 \, \text{kJ of energy released} \][/tex]
None of the provided options (13.5 kJ of energy released, 13.5 kJ of energy absorbed, 84.6 kJ of energy released, 84.6 kJ of energy absorbed) match exactly 42.3 kJ of energy released, but given the calculated value, our confidence is that:
\[ \boxed{-42.3 \, \text{kJ of energy released}}
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