Chemical Thermodynamics Question 216
Question: From given following equations and $ \Delta H{}^\circ $ values, determine the enthalpy of reaction at 298 K for the reaction:
$ C_2H_4( g )+6F_2( g )\xrightarrow{{}}2CF_4( g )+4HF( g ) $
$ H_2(g)+F_2(g)\xrightarrow{{}}2HF(g);\Delta H_1^{{}^\circ }=-537kJ $
$ C( s )+2F_2( g )\xrightarrow{{}}CF_4( g );\Delta H_2^{{}^\circ }=-680kJ $
$ 2C( s )+2H_2( g )\xrightarrow{{}}C_2H_4( g );\Delta H_3^{{}^\circ }=52kJ $
Options:
A) $ ~-1165 $
B) $ -2486 $
C) $ +1165 $
D) $ +2486 $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ \Delta {{H}^{{}^\circ }}=2\times \Delta H_1^{{}^\circ }+2\times \Delta H_2^{{}^\circ }-\Delta H_3^{{}^\circ }=-2486,kJ $
BETA
