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The equilibrium constant for the reaction $SO_2 (g)+\large\frac{1}{2}$$ O_2 (g) \rightleftharpoons SO_3(g)$ is $5 \times 10 ^{-2}\;atm$ The equilibrium consant of the reaction $250_3(g) \rightleftharpoons 2SO_2(g) +O_2 (g) $ would be

\[\begin {array} {1 1} (a)\;100\;atm & \quad (b)\;200\;atm \\ (c)\;4 \times 10^2\;atm & \quad (d)\;6.25 \times 10^4 \;atm \end {array}\]

1 Answer

$(d) 6.25 \times 10^{4}\;atm$
answered Nov 7, 2013 by pady_1
 

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