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another question guys! :)
- Thread starter JustwantDDS
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K for the first reaction = (NO2)^2 / [(NO)^2 * O2] = 2
K for the second reaction = [O2]^(1/2) * NO / NO2
Take the square root of both sides for the first equation and you get:
NO2/[NO * (O2)^1/2] = 2^(1/2)
Take the reciprocal of both sides and you get K for the second reaction which would be 1/[2^(1/2)]
Since 1 is equivalent to 1^(1/2), 1/[2^(1/2)] = (1/2.0)^(1/2).
Therefore, the answer is A.
K for the second reaction = [O2]^(1/2) * NO / NO2
Take the square root of both sides for the first equation and you get:
NO2/[NO * (O2)^1/2] = 2^(1/2)
Take the reciprocal of both sides and you get K for the second reaction which would be 1/[2^(1/2)]
Since 1 is equivalent to 1^(1/2), 1/[2^(1/2)] = (1/2.0)^(1/2).
Therefore, the answer is A.
This is easy. Destroyer has a problem just like this.
The new reaction has been flipped, so put 1 over your K value (K=2.0) --> 1/2.0
Then, since you cut the # of moles in half, you square root it (aka to the 1/2 power) --- > (1/2.0)^1/2
If the # of moles doubled, you would have squared your 1/2.0 instead of square rooting it.
The new reaction has been flipped, so put 1 over your K value (K=2.0) --> 1/2.0
Then, since you cut the # of moles in half, you square root it (aka to the 1/2 power) --- > (1/2.0)^1/2
If the # of moles doubled, you would have squared your 1/2.0 instead of square rooting it.
