Given the following thermochemical equations: | vedclass

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(B) According to Hess's Law,the enthalpy change of a reaction is the same whether it occurs in one step or several steps.To obtain the target reaction

Vedclass · Accepted Answer

$450$

Vedclass · Answer

(B) According to Hess's Law,the enthalpy change of a reaction is the same whether it occurs in one step or several steps.To obtain the target reaction $(4) \ C_{(s)} + O_{2_{(g)}} 	o CO_{2_{(g)}}$,we can add equations $(1)$,$(2)$,and $(3)$:$(1) \ H_2O_{(g)} + C_{(s)} 	o CO_{(g)} + H_{2(g)} ; \Delta H_1 = 100 \ kJ$$(2) \ CO_{(g)} + \frac{1}{2}O_{2_{(g)}} 	o CO_{2_{(g)}} ; \Delta H_2 = -300 \ kJ$$(3) \ H_{2(g)} + \frac{1}{2}O_{2_{(g)}} 	o H_2O_{(g)} ; \Delta H_3 = -250 \ kJ$Adding $(1) + (2) + (3)$:$(H_2O_{(g)} + C_{(s)} + CO_{(g)} + \frac{1}{2}O_{2_{(g)}} + H_{2(g)} + \frac{1}{2}O_{2_{(g)}}) 	o (CO_{(g)} + H_{2(g)} + CO_{2_{(g)}} + H_2O_{(g)})$Canceling common terms on both sides:$C_{(s)} + O_{2_{(g)}} 	o CO_{2_{(g)}}$The total enthalpy change is $\Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 = 100 + (-300) + (-250) = -450 \ kJ$.Since $\Delta H_4 = -x \ kJ$,we have $-x = -450$,which implies $x = 450$.

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

Similar Questions

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

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

$C-H$ $413$

$C-C$ $348$

$C=C$ $610$

$H-H$ $436$

What is enthalpy of reaction?

The bond enthalpies of $C-C, C=C, H-H$ and $C-H$ bonds are $360, 600, 400$ and $410 \ kJ \ mol^{-1}$ respectively. What is the heat of hydrogenation of ethylene?

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