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An excess of KO(2) is placed in a closed...

An excess of `KO_(2)` is placed in a closed container of `CO_(2)`(g). After reaction is completed. Will the gas pressure be same, greater or less than initial value. Example.

Text Solution

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`2KO_(2(s))+CO_(2(g))toK_(2)CO_(3(s))+(3)/(2)O_(2(g))`
the gas pressure will increase as number of moles of gaseous product is 3/2 while that of gaseous reactant was one.
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