A) 6.3 atm
B) 6.93 atm
C) 0.63 atm
D) 0.693 atm
Correct Answer: B
Solution :
\[C\left( s \right)+C{{O}_{2}}\left( g \right)\xrightarrow[{}]{}2CO\left( g \right)\] |
Apply law of mass action, |
\[{{K}_{P}}=\frac{{{\left( {{P}_{C{{O}_{2}}}} \right)}^{2}}}{{{P}_{C{{O}_{2}}}}}\] |
Given: \[{{\operatorname{K}}_{P}}=65\operatorname{and}\,{{P}_{CO}}=10{{P}_{C{{O}_{2}}}}\] |
\[63=\frac{{{\left( 10{{P}_{C{{O}_{2}}}} \right)}^{2}}}{{{P}_{C{{O}_{2}}}}}\] |
Or \[63=\frac{100{{\left( {{P}_{C{{O}_{2}}}} \right)}^{2}}}{{{P}_{C{{O}_{2}}}}}\operatorname{or}\,63=100{{P}_{C{{O}_{2}}}}\] |
\[{{\operatorname{P}}_{C{{O}_{2}}}}=\frac{63}{100}=0.63atm\] |
\[{{\operatorname{P}}_{C{{O}_{2}}}}=102{{P}_{C{{O}_{2}}}}=10\times 0.63=6.3atm\] |
\[{{\operatorname{P}}_{total}}={{P}_{C{{O}_{2}}}}+{{P}_{CO}}=0.63+6.3=6.93atm\] |
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