| (A)\[Ca{{C}_{2}}(s)+2{{H}_{2}}O(l)\xrightarrow{{}}Ca{{(OH)}_{2}}(s)\] \[+\,{{C}_{2}}{{H}_{2}}(g)D{{H}^{o}}(kJ\,mo{{l}^{-1}})-127.9\] |
| (B) \[Ca(s)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}CaO(s)-635.1\] |
| (C) \[CaO(s)+{{H}_{2}}O(I)\xrightarrow{{}}Ca{{(OH)}_{2}}(s)-65.2\] |
| (D) \[C(s)+{{O}_{2}}(s)\xrightarrow{{}}C{{O}_{2}}(s)-393.5\] |
| (E) \[{{C}_{2}}{{H}_{2}}(g)+\frac{5}{2}{{O}_{2}}(g)\xrightarrow{{}}2C{{O}_{2}}(g)\] \[+\,{{H}_{2}}O(e)-1299.58\] |
A) \[-\text{ }59.82\text{ }KJ\text{ }mo{{l}^{-1}}\]
B) \[~+\text{ }59.82\text{ }KJ\text{ }mo{{l}^{-1}}\]
C) \[-190.22\text{ }KJ\text{ }mo{{l}^{-1}}\]
D) \[+\,190.22\,KJ\,mo{{l}^{-1}}\]
Correct Answer: A
Solution :
We have to calculate \[\Delta {{H}^{o}}\] for the following reaction in which one mole of \[Ca{{C}_{2}}(S)\]is formed form its elements. \[Ca(S)+2C(S)\xrightarrow{{}}Ca{{C}_{2}}(S)\] This is obtained when we operate. \[(B)+(C)+z(D)-(E)-a\] \[\therefore \] \[\Delta H=(-635.1)+(-65.2)+2(-393.5)\] \[-(-1299.58)-(-127.9)\,kJ\,mo{{l}^{-}}\] \[=-59.82\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}\]
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