A) \[\text{800 kJ mo}{{\text{l}}^{\text{--1}}}\]
B) \[\text{100 kJ mo}{{\text{l}}^{\text{--1}}}\]
C) \[\text{200 kJ mo}{{\text{l}}^{\text{--1}}}\]
D) \[\text{400 kJ mo}{{\text{l}}^{\text{--1}}}\]
Correct Answer: A
Solution :
The reaction for \[{{\Delta }_{f}}\text{H }\!\!{}^\circ\!\!\text{ (XY)}\] \[\frac{1}{2}{{X}_{2}}(g)+\frac{1}{2}{{Y}_{2}}(g)\xrightarrow{{}}XY(g)\] Bond energies of \[{{X}_{2}},{{Y}_{2}}\] and \[XY\] are \[X,\frac{X}{2},X\] respectively \[\therefore \] \[\Delta H=\left( \frac{X}{2}+\frac{X}{4} \right)-x=-200\] On solving, we get\[\Rightarrow -\frac{X}{2}+\frac{X}{4}=-200\] |
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