| \[\text{ }\!\!\Delta\!\!\text{ H}\,\text{(kJ/}\,\text{mol)}\] | |
| \[1/2A\to B\] | + 150 |
| \[3B\to 2C+D\] | - 125 |
| \[E+A\to 2D\] | + 350 |
A) 525 kJ/mol
B) - 175 kJ/mol
C) -325 kJ/mol
D) 325 kJ/mol
Correct Answer: B
Solution :
\[\frac{1}{2}A\xrightarrow[{}]{{}}B;\] \[\Delta H=150\,kJ/mol\] ?(i) \[3B\xrightarrow[{}]{{}}2C+D;\Delta H=-125\,kJ/mol\] ?(ii) \[\underline{E+A\xrightarrow[{}]{{}}2D;\Delta H=+350\,kJ/mol}\] ?(iii) By\[[2\times (i)+(ii)]-(iii),\]we have \[B+D\xrightarrow[{}]{{}}E+2C\] \[\therefore \] \[\Delta H=150\times 2+(-125)-350\] \[=-175\,kJ/mol.\]
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