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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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Knowledge Check

  • When excess potassium super oxide is placed in a container containing CO_(2) , then

    A
    The pressure of the container decreases
    B
    The pressure of the container increases
    C
    The pressure of the container remains constant
    D
    The pressure of the container first increases and then decreases

  • For the reaction, A(g)+2B(g)rarrC(g)+D(g) (dx)/(dt)=k[A][B]^(-2) Initial pressure of A and B are respectively 0.60 atm and 0.80 atm. At a time when pressure of C is 0.20 atm, rate of the reaction, realative of the initial value is

    A
    44202
    B
    `1/(48)`
    C
    44200
    D
    `1/(24)`

  • For the reaction, A(g)+2B(g) to C(g)+D(g) (dx)/(dt)=k[A][B]^(2) Initial pressure of A and B are respectively 0.60 atm and 0.80 atm. At a time when pressure of C is 0.20 atm, rate of the reaction, relative to the initial value is

    A
    `(1)/(6)`
    B
    `(1)/(48)`
    C
    `(1)/(4)`
    D
    `(1)/(24)`