| \[3Fe\left( s \right)+4{{H}_{2}}O\left( g \right).=F{{e}_{3}}{{O}_{4}}, \left( s \right)+4{{H}_{2}} \left( g \right)\]is equal: |
A) \[{{\left( P{{H}_{2}} \right)}^{4}}\left( PE{{e}_{3}}{{O}_{4}} \right)\]
B) \[\frac{P{{H}_{2}}}{P{{H}_{2}}O}\]
C) \[\frac{{{\left( P{{H}_{2}} \right)}^{4}}}{{{\left( P{{H}_{2}}O \right)}^{4}}}\]
D) \[\frac{\left( P{{H}_{2}} \right)\left( P\,F{{e}_{3}}\,{{O}_{4}} \right)}{P\,Fe}\]
Correct Answer: C
Solution :
\[{{\operatorname{K}}_{p}}\]is partial pressure constant For general reaction \[\operatorname{aA}+bB\rightleftharpoons xX+yY\] \[{{\operatorname{K}}_{p}}=\frac{{{P}^{x}}_{X}{{P}^{y}}_{Y}}{{{P}^{a}}A\,{{P}^{b}}B}\] \[{{\operatorname{K}}_{p}}~is dimensionless\]
You need to login to perform this action.
You will be redirected in
3 sec
