A) \[-4.0\,kcal\]
B) \[+40\,kcal\]
C) \[-80\,kcal\]
D) \[+80\,kcal\]
Correct Answer: C
Solution :
[c] Given, |
(i) \[C{{H}_{4}}+2{{O}_{2}}\to C{{O}_{2}}+2{{H}_{2}}O\], \[\Delta H=+20\]kcal |
(ii) \[C+{{O}_{2}}\to C{{O}_{2}}\] , \[\Delta H=-40\,\,kcal.\] |
(iii) \[{{H}_{2}}+\frac{1}{2}{{O}_{2}}\to {{H}_{2}}O\], \[\Delta H=-10\,\,kcal.\] |
Aim: \[C+2{{H}_{2}}\to C{{H}_{4}}\] |
(ii) + 2 × (iii) - (i) gives. |
\[\Delta H=-40+2\,(-10)-(+\,20)=-80\, & \,kcal.\] |
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