| \[CaC{{O}_{3}}(s)\,\,\rightleftharpoons \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {{H}^{0}}=42.8\,\,kcals\] |
| \[CaO\left( s \right)+C{{O}_{2}}\left( g \right)\] |
| \[CaO(s)+3C(s)\,\,\rightleftharpoons \,\,\,\,\,\,\,\,\,\,\,\,\Delta {{H}^{0}}=111\,\,kcals\] |
| \[Ca{{C}_{2}}+\text{ }CO\left( g \right)\] |
A) 102.6 kcals
B) 221.78 kcals
C) 307.6 kcals
D) 453.46 kcals
Correct Answer: C
Solution :
\[CaC{{O}_{3}}(s)\,\,\rightleftharpoons \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {{H}^{0}}=42.8\,\,kcals\] \[CaO\left( s \right)\,\,+\,\,C{{O}_{2}}\left( g \right)\] \[\operatorname{CaO}\left( s \right) \,+\, 3C\left( s \right)\,\,\,\rightleftharpoons \,\,\,\,\,\,\,\,\,\,\,\,\,\Delta {{H}^{0}}=111\,\,kcals\] \[Ca{{C}_{2}}\left( s \right)+CO\left( g \right)\] \[{{\operatorname{CaCO}}_{3}}\left( s \right) + 3C\left( s \right)\,\,\rightleftharpoons \] 0 \[Ca{{C}_{2}}(s)+CO(g)+C{{O}_{2}}(g)\,\,\Delta H\,\,=\,153.8\,\,kcals\] Thus heat required to prepare 1 mole of \[Ca{{C}_{2}}\] from \[\operatorname{CaC}{{O}_{3}}= 153.8 kcal\] Molecular weight of \[{{\operatorname{CaC}}_{2}} = 40 + 24 = 64\] 64 g of \[Ca{{C}_{2}}\] requires 153.8 kcals of heat 128 g of \[Ca{{C}_{2}}\] requires 307.6 kcals of heat
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