1.

For a reaction NO_(g) +O_(2) (g)rarr 2NO_(2)(g) Rate=k[NO^(2)] [O_(2)] if the volume of the reaction vessel is doubled the rate of the reaction :

Answer»

will diminish to 1/4 of initial value
will diminish to 1/8 of initial value will grow 4 times
will grow 4times
will grow 8 times

SOLUTION :(B) On increasing the volume to twice the value the concentration of each species is REDUCED by a factor of 2 . Therefore
RATE = `k[NO]^(2) [O_(2)]`
`"Rate"_(2)= k(([NO])/(2))^(2)(([O_(2)])/(2))`
`("Rate"_(2))/("Rate"_(1)) = (1)/(8)`


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