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Since the reaction is first order with respect to H2, if the concentration of H2 were halved, the rate of the reaction would be halved. This can be seen by entering one for each value in the rate equation, then changing the value of [H2] to 1/2 while keeping the other values the same: The rate changes from 1 to 1/2.
It will decrease by half.
The rate would be four times larger. Impossible.
zff
Substrate concentration is directly proportional to rate of reaction because more enzymes are able to act on it but this happen only until all the active sites are covered than reaction rate becomes constant.
Since the reaction is first order with respect to H2, if the concentration of H2 were halved, the rate of the reaction would be halved. This can be seen by entering one for each value in the rate equation, then changing the value of [H2] to 1/2 while keeping the other values the same: The rate changes from 1 to 1/2.
It will decrease by half.
This rate law suggests its rate to be direct proportional to H2 concentration, thus halving this would implicate a halved reaction rate.
The rate would also behalved
The rate would be four times larger. Impossible.
The rate would be four times larger. Impossible.
The reaction rate would decrease
It will also be halved as circumference = 2Ï€r
zff
Substrate concentration is directly proportional to rate of reaction because more enzymes are able to act on it but this happen only until all the active sites are covered than reaction rate becomes constant.
The rate would quadruple (increase by a factor of 4). This is because the rate depends on the SQUARE of the concentration of NO.
It will be a quarter of what it was.