| \[\mathbf{C+O2}\to \mathbf{C}{{\mathbf{O}}_{\mathbf{2}}}~\,\,\,\,\,\,\,\,\,,\Delta \mathbf{H}{}^\circ =-\mathbf{xKJ}\] |
| \[\mathbf{2CO}+{{\mathbf{O}}_{\mathbf{2}}}\to \mathbf{2C}{{\mathbf{O}}_{\mathbf{2}}}\mathbf{ }\,\,\,\,\mathbf{,}\Delta \mathbf{H{}^\circ =}-\mathbf{yKJ}\] |
| The enthalpy of carbon monoxide will be: |
A) \[\operatorname{y}-2x\]
B) \[\frac{2x-\operatorname{y}}{2}\]
C) \[\frac{\operatorname{y}-2x}{2}\]
D) \[2x-y\]
Correct Answer: C
Solution :
\[\frac{y-2x}{2}\] It we subtract both the equations, We get the result, i.e \[\operatorname{C}\left( s \right)+\frac{1}{2}{{O}_{2}} \left( g \right)\to CO\left( g \right)\] Standard enthalpy of formation is \[\Sigma \Delta f H{}^\circ \left( products \right) - \Sigma \Delta f H{}^\circ \left( reactants \right)\] \[\operatorname{Hence}, we get\frac{y-2X}{2}\]
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