With the help of the following data,find out | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The reaction is: $C_2H_{4(g)} + H_{2(g)} 	o C_2H_{6(g)}$Change in enthalpy $(\Delta H)$ is calculated as: $\Delta H = \sum 	ext{Bond energy of r

Vedclass · Accepted Answer

$-128$

Vedclass · Answer

(A) The reaction is: $C_2H_{4(g)} + H_{2(g)} 	o C_2H_{6(g)}$Change in enthalpy $(\Delta H)$ is calculated as: $\Delta H = \sum 	ext{Bond energy of reactants} - \sum 	ext{Bond energy of products}$Reactants:$1 	imes (C=C) = 610 \ kJ$$4 	imes (C-H) = 4 	imes 413 = 1652 \ kJ$$1 	imes (H-H) = 436 \ kJ$Total reactant bond energy = $610 + 1652 + 436 = 2698 \ kJ$Products:$1 	imes (C-C) = 348 \ kJ$$6 	imes (C-H) = 6 	imes 413 = 2478 \ kJ$Total product bond energy = $348 + 2478 = 2826 \ kJ$$\Delta H = 2698 - 2826 = -128 \ kJ$Thus,the change in heat content is $-128 \ kJ$.

Bond	Bond energy $(kJ \ mol^{-1})$
$C-H$	$413$
$C-C$	$348$
$C=C$	$610$
$H-H$	$436$

With the help of the following data,find out the change in heat content for the reaction in $kJ$:
$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$

Bond Bond energy $(kJ \ mol^{-1})$

$C-H$ $413$

$C-C$ $348$

$C=C$ $610$

$H-H$ $436$

Similar Questions

Calculate the heat of formation of ethyl alcohol in $kcal/mol$ from the following data:
$C_{(s)} + O_{2(g)} \to CO_{2(g)}, \Delta H = -94 \ kcal$
$H_{2(g)} + \frac{1}{2} O_{2(g)} \to H_2O_{(l)}, \Delta H = -68 \ kcal$
$C_2H_5OH_{(l)} + 3O_{2(g)} \to 2CO_{2(g)} + 3H_2O_{(l)}, \Delta H = -327 \ kcal$

Given that the molar combustion enthalpies of benzene,cyclohexane,and hydrogen are $x, y$,and $z$ respectively,the molar enthalpy of hydrogenation of benzene to cyclohexane is

What is the heat of formation of $HCl_{(g)}$ from the following equation (in $kJ$)?
$H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)} \quad \Delta_{r}H = -194 \ kJ$

Average bond enthalpy of water is $464.5 \text{ kJ mol}^{-1}$. If the energy required to break the first $O-H$ bond is $502 \text{ kJ mol}^{-1}$,how much energy per mol is required to break the second $O-H$ bond?

$C_{(diamond)} + O_2 \to CO_2; \Delta H = -395.3 \ kJ/mole$
$C_{(graphite)} + O_2 \to CO_2; \Delta H = -393.4 \ kJ/mole$
$C_{(graphite)} \to C_{(diamond)}; \Delta H = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

With the help of the following data,find out the change in heat content for the reaction in $kJ$:$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$ Bond Bond energy $(kJ \ mol^{-1})$ $C-H$ $413$ $C-C$ $348$ $C=C$ $610$ $H-H$ $436$