1. Tardigrade
  2. Question
  3. Chemistry
  4. From given following equations and ΔH° values at 298K, determine the enthalpy of reaction (in kJ.:- C2H4(.g.)+6F2(.g.) arrow 2(CF)4(.g.)+4HF(.g.);(ΔH)o= ? H2(.g.)+F2(.g.) arrow 2HF(.g.);(ΔH)1° =-537kJ C(. s .)+2F2(.g.) arrow (CF)4(.g.);(ΔH)2° =-680kJ 2C(. s .)+2H2(.g.) arrow C2H4(.g.);ΔH30=52kJ

Q. From given following equations and $ΔH^\circ $ values at $298K,$ determine the enthalpy of reaction (in $kJ\left.:-$
$C_{2}H_{4}\left(\right.g\left.\right)+6F_{2}\left(\right.g\left.\right) \rightarrow 2\left(CF\right)_{4}\left(\right.g\left.\right)+4HF\left(\right.g\left.\right);\left(ΔH\right)^{o}=$ ?
$H_{2}\left(\right.g\left.\right)+F_{2}\left(\right.g\left.\right) \rightarrow 2HF\left(\right.g\left.\right);\left(ΔH\right)_{1}^\circ =-537kJ$
$C_{\left(\right. s \left.\right)}+2F_{2}\left(\right.g\left.\right) \rightarrow \left(CF\right)_{4}\left(\right.g\left.\right);\left(ΔH\right)_{2}^\circ =-680kJ$
$2C_{\left(\right. s \left.\right)}+2H_{2}\left(\right.g\left.\right) \rightarrow C_{2}H_{4}\left(\right.g\left.\right);ΔH_{3}^{0}=52kJ$

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Solution:

$ΔH^\circ =2\times ΔH_{1}^{0}+2\times ΔH_{2}^{0}-ΔH_{3}^\circ $
$=2\times \left(\right.-537\left.\right)+2\times \left(\right.-680\left.\right)-52$
$=-2486kJ$

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