Answer :
To determine the net change in enthalpy (∆H) for the formation of one mole of nitric oxide (NO) from nitrogen, hydrogen, and oxygen, we need to consider the enthalpy changes of the given reactions:
### Step 1: Formation of Ammonia
The first step involves the formation of ammonia from nitrogen and hydrogen:
[tex]\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \][/tex]
The enthalpy change for this reaction is given as:
[tex]\[ \Delta H = -92 \, \text{kJ} \][/tex]
This means that producing 2 moles of NH₃ releases 92 kJ of energy.
### Step 2: Formation of Nitric Oxide and Water
The second step involves the formation of nitric oxide and water from ammonia and oxygen:
[tex]\[ 4 NH_3(g) + 5 O_2(g) \rightarrow 4 NO(g) + 6 H_2O(g) \][/tex]
The enthalpy change for this reaction is given as:
[tex]\[ \Delta H = -905 \, \text{kJ} \][/tex]
This means that producing 4 moles of NO releases 905 kJ of energy.
### Calculation of Enthalpy Change per Mole of NO
To find the net change in enthalpy for the formation of one mole of NO, we need to normalize these values to per mole of NO produced.
#### Step 1: Contribution of the First Reaction
The first reaction produces 2 moles of NH₃ with an enthalpy change of -92 kJ. We typically need to link it to NO production.
To match the amount of NO, the second reaction which involves 4 moles of NH₃ matches the second reaction’s enthalpy:
[tex]\[ \Delta H = -92 \, \text{kJ} \text{ per 2 moles of } NH_3 \][/tex]
[tex]\[ \text{For 1 mole of } NH_3: \frac{-92 \, \text{kJ}}{2 \, \text{moles}} = -46 \, \text{kJ/mole NH_3} \][/tex]
Since the second reaction needs 4 moles of NH₃:
[tex]\[ 4 \times (-46 \, \text{kJ/mole NH_3}) = -184 \, \text{kJ for 4 moles NH_3} \][/tex]
Net energy released for producing matching NO:
[tex]\[ \Delta H = -92 \, \text{kJ per 2 moles (hence per 4 NH₃)} = 2 \times (-46(\text{kJ/mole NH_3}) = -92 \, \text{kJ}\][/tex]
Effectively:
[tex]\[ -92 \div 2 = -46 \, \text{kJ} \text{ per NO via 4 NH₃} \text{ duly balances for NO required} \][/tex]
#### Step 2: Contribution of the Second Reaction
The second reaction directly results in the production of 4 moles of NO with an enthalpy change of:
[tex]\[ \Delta H = -905 \, \text{kJ} \][/tex]
We want to find the enthalpy change per one mole of NO:
[tex]\[ \text{Per one mole of } NO: \frac{-905 \, \text{kJ}}{4 \, \text{moles}} = -226.25 \, \text{kJ/mole NO} \][/tex]
### Net Enthalpy Change for One Mole of NO
The total enthalpy change per mole of NO is calculated by summing the contributions from both reactions:
[tex]\[ \Delta H (\text{per mole of NO}) = -46 \, \text{kJ (first reaction)} + -226.25 \, \text{kJ (second reaction)} = -272.25 \, \text{kJ} \][/tex]
### Rounding to the Nearest kJ
Finally, rounding the computed value to the nearest kJ, we get:
[tex]\[ \Delta H (\text{net per mole of NO}) \approx -272 \, \text{kJ} \][/tex]
Thus, the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen, and oxygen is [tex]\( \boxed{-272 \, \text{kJ}} \)[/tex].
### Step 1: Formation of Ammonia
The first step involves the formation of ammonia from nitrogen and hydrogen:
[tex]\[ N_2(g) + 3 H_2(g) \rightarrow 2 NH_3(g) \][/tex]
The enthalpy change for this reaction is given as:
[tex]\[ \Delta H = -92 \, \text{kJ} \][/tex]
This means that producing 2 moles of NH₃ releases 92 kJ of energy.
### Step 2: Formation of Nitric Oxide and Water
The second step involves the formation of nitric oxide and water from ammonia and oxygen:
[tex]\[ 4 NH_3(g) + 5 O_2(g) \rightarrow 4 NO(g) + 6 H_2O(g) \][/tex]
The enthalpy change for this reaction is given as:
[tex]\[ \Delta H = -905 \, \text{kJ} \][/tex]
This means that producing 4 moles of NO releases 905 kJ of energy.
### Calculation of Enthalpy Change per Mole of NO
To find the net change in enthalpy for the formation of one mole of NO, we need to normalize these values to per mole of NO produced.
#### Step 1: Contribution of the First Reaction
The first reaction produces 2 moles of NH₃ with an enthalpy change of -92 kJ. We typically need to link it to NO production.
To match the amount of NO, the second reaction which involves 4 moles of NH₃ matches the second reaction’s enthalpy:
[tex]\[ \Delta H = -92 \, \text{kJ} \text{ per 2 moles of } NH_3 \][/tex]
[tex]\[ \text{For 1 mole of } NH_3: \frac{-92 \, \text{kJ}}{2 \, \text{moles}} = -46 \, \text{kJ/mole NH_3} \][/tex]
Since the second reaction needs 4 moles of NH₃:
[tex]\[ 4 \times (-46 \, \text{kJ/mole NH_3}) = -184 \, \text{kJ for 4 moles NH_3} \][/tex]
Net energy released for producing matching NO:
[tex]\[ \Delta H = -92 \, \text{kJ per 2 moles (hence per 4 NH₃)} = 2 \times (-46(\text{kJ/mole NH_3}) = -92 \, \text{kJ}\][/tex]
Effectively:
[tex]\[ -92 \div 2 = -46 \, \text{kJ} \text{ per NO via 4 NH₃} \text{ duly balances for NO required} \][/tex]
#### Step 2: Contribution of the Second Reaction
The second reaction directly results in the production of 4 moles of NO with an enthalpy change of:
[tex]\[ \Delta H = -905 \, \text{kJ} \][/tex]
We want to find the enthalpy change per one mole of NO:
[tex]\[ \text{Per one mole of } NO: \frac{-905 \, \text{kJ}}{4 \, \text{moles}} = -226.25 \, \text{kJ/mole NO} \][/tex]
### Net Enthalpy Change for One Mole of NO
The total enthalpy change per mole of NO is calculated by summing the contributions from both reactions:
[tex]\[ \Delta H (\text{per mole of NO}) = -46 \, \text{kJ (first reaction)} + -226.25 \, \text{kJ (second reaction)} = -272.25 \, \text{kJ} \][/tex]
### Rounding to the Nearest kJ
Finally, rounding the computed value to the nearest kJ, we get:
[tex]\[ \Delta H (\text{net per mole of NO}) \approx -272 \, \text{kJ} \][/tex]
Thus, the net change in enthalpy for the formation of one mole of nitric oxide from nitrogen, hydrogen, and oxygen is [tex]\( \boxed{-272 \, \text{kJ}} \)[/tex].
