Respuesta :
Answer: The [tex]\Delta H^o_{formation}[/tex] for the reaction is 591.9 kJ.
Explanation:
Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.
According to this law, the chemical equation is treated as ordinary algebraic expressions and can be added or subtracted to yield the required equation. This means that the enthalpy change of the overall reaction is equal to the sum of the enthalpy changes of the intermediate reactions.
The chemical reaction for the formation reaction of [tex]N_2O[/tex] is:
[tex]N_2(g)+\frac{1}{2}O_2(g)\rightarrow N_2O(g)[/tex] [tex]\Delta H^o_{formation}=?[/tex]
The intermediate balanced chemical reaction are:
(1) [tex]2NH_3(g)+3N_2O(g)\rightarrow 4N_2(g)+3H_2O(l)[/tex] [tex]\Delta H_1=-1010kJ[/tex] ( ÷ 3)
(2) [tex]4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(l) [tex]\Delta H_2=1531kJ[/tex] ( ÷ 6)
Reversing Equation 1 and then adding both the equations, we get the enthalpy change for the chemical reaction.
[tex]\Delta H^o_{formation}=[\frac{\Delta H_1}{3}]+[\frac{\Delta H_2}{6}][/tex]
Putting values in above equation, we get:
[tex]\Delta H^o_{formation}=[\frac{1010}{3}]+[\frac{1531}{6}]\\\\\Delta H^o_{formation}=591.9kJ[/tex]
Hence, the [tex]\Delta H^o_{formation}[/tex] for the reaction is 591.9 kJ.
Hess's law is defined as the sum of amount of heat absorbed or released in the given chemical equation remains constant, irrespective of the steps involved in the reaction. The [tex]\Delta \text H^0_{\text{formation}}&=591.9 \text{kJ}[/tex] is for the given reaction.
Given that,
- [tex]\Delta \text H^0_{1}&=-1010 \text{kJ}[/tex]
- [tex]\Delta \text H^0_{2}&=-1531 \text{kJ}[/tex]
- [tex]\Delta \text H^0_{\text{formation}}&=? \text{kJ}[/tex]
Now, the given chemical equations are:
- [tex]\text{N}_2_{\text (g)} + \dfrac{1}{2}\text O_2_{(\text g)} \rightarrow \text N_2\text O\;\;\;\;\Delta \text H^0_{\text{formation}}&=? \text{kJ}[/tex]
The intermediate reactions between the above equation are:
- [tex]\text {2 NH}_3_\text{(g)} + 3\text N_2\text O \rightarrow 4\text N_2 +3\text H_2\text O\;\;\;\;\;\Delta \text H^0_{1}&=-1010 \text{kJ}[/tex]
- [tex]\text {4 NH}_3_\text{(g)} + \text O_2 \rightarrow 2\text N_2 +6\text H_2\text O\;\;\;\;\;\Delta \text H^0_{1}&=-1531\text{kJ}[/tex]
Reversing the equation and then adding both the values, the enthalpy change becomes:
- [tex]\Delta \text H^0_{\text{formation}}&=\dfrac{\Delta \text {H}_1}{3}+\dfrac{\Delta \text {H}_2}{6}\\\\\Delta \text H^0_{\text{formation}}&=\dfrac{1010}{3}+\dfrac{1531}{6}\\\\\Delta \text H^0_{\text{formation}}&= 591.9 \text{kJ}[/tex]
Therefore, the enthalpy change of the reaction is [tex]\Delta \text H^0_{\text{formation}}&=591.9 \text{kJ}[/tex].
To know more about enthalpy change, refer to the following link:
https://brainly.com/question/20629950?referrer=searchResults
