Respuesta :
Answer:
The answer is "Option b".
Explanation:
Given equation and value:
[tex]\Rightarrow C_2H_2 \ (g)+ \frac{5}{2}O_2 \ (g) \longrightarrow 2CO_2\ (g)+H_2O\ (g)[/tex]
[tex]\left\begin{array}{ccc}C \equiv C&839\\C-H &413\\O=O &495\\ C=O &799\\ O-H &467 \end{array}\right \\\\[/tex]
calculating equation value:
[tex]\Rightarrow 839 + 2(413) +\frac{5}{2} (495) \longrightarrow 2(2)799+2 \times 467 +E[/tex]
[tex]\Rightarrow 839 + 826 +1237.5 = 3196+934 +E\\\\\Rightarrow 2902.5 =4130 +E\\\\\Rightarrow 2902 = 4130 +E\\\\\Rightarrow 2902 - 4130 +E\\\\\Rightarrow E = -1228 \ KJ \\\\[/tex]
The heat(enthalpy) of combustion of acetylene = -1228 kJ
The heat of combustion refers to the amount of heat released when 1 mole of a substance is burned.
The chemical reaction is given in the equation;
- [tex]\mathbf{C_2H_{2(g)} + \dfrac{5}{2}O_{2(g)} \to 2CO_{2(g)}+H_2O_{(g)}}[/tex]
The bond energy of the reactant is:
- [tex]\mathbf{=( 1 \times C \equiv C) + ( \dfrac{5}{2} O =O)+(2 \times (C-H))}[/tex] Â
Following the bond energies given in the question, we have:
= ( 1 × 839) +(5/2 × 495) +(2 × 413)
= 2902.5 kJ
The bond energy of the product is:
- [tex]\mathbf{4 \times C=O ) + (2 \times O-H)}[/tex]
= (4× 799) +(2 × 467)
= (3196 + 934) kJ
= 4130 kJ
The heat(enthalpy) of combustion of acetylene = bond energy of reactant - bond energy of the product.
The heat(enthalpy) of combustion of acetylene = 2902.5 kJ - 4130 kJ
The heat(enthalpy) of combustion of acetylene = -1227.5 kJ
≅  -1228 kJ
Learn more about heat of combustion here:
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