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.
In the rate law given as rate = k[NO2][H2], the concentration of NO does not appear, so the rate of the reaction is independent of its concentration. Therefore, if the concentration of NO were halved, it would have no effect on the rate of the reaction. The reaction rate would remain unchanged as long as the concentrations of NO2 and H2 remain constant.
The rate would be four times larger. Impossible.
In the rate law given as rate = k[NO2][H2], the reaction rate is directly proportional to the concentration of both NO2 and H2. If the concentration of H2 is halved, the reaction rate would also be halved, assuming the concentration of NO2 remains constant. This is because the rate depends linearly on the concentration of H2, so any decrease in H2 concentration results in a proportional decrease in the overall reaction rate.
In the given rate law, the rate of the reaction is dependent on the concentration of NO and possibly other reactants. If the concentration of NO is halved, the rate of the reaction would decrease proportionally, assuming that NO is a reactant in the rate law. Specifically, if the rate law is of the form rate = k[NO]^n[other species], the rate would be affected by the new concentration of NO, resulting in a reduced reaction rate. The exact impact on the rate would depend on the order of the reaction with respect to NO.
If the concentration of H2 is halved, the rate of the reaction will also be halved. This is because the rate of a reaction is directly proportional to the concentration of reactants in the rate law equation. Thus, reducing the concentration of H2 will directly impact the rate of the reaction.
If the concentration of NO is halved, the rate of the reaction will also be halved. This is because the rate of the reaction is directly proportional to the concentration of NO raised to the power of its coefficient in the rate law (in this case 1). So, halving the concentration of NO will result in a proportional decrease in the rate of the reaction.
It will decrease by half.
It will also be halved as circumference = 2Ï€r
In the rate law given as rate = k[NO2][H2], the concentration of NO does not appear, so the rate of the reaction is independent of its concentration. Therefore, if the concentration of NO were halved, it would have no effect on the rate of the reaction. The reaction rate would remain unchanged as long as the concentrations of NO2 and H2 remain constant.
The rate would be four times larger. Impossible.
If the concentration of H2 is halved, it would also halve the rate of the reaction, assuming H2 is a reactant in the rate law. This is because the rate law is directly proportional to the concentrations of reactants.
In the given rate law, the rate of the reaction is dependent on the concentration of NO and possibly other reactants. If the concentration of NO is halved, the rate of the reaction would decrease proportionally, assuming that NO is a reactant in the rate law. Specifically, if the rate law is of the form rate = k[NO]^n[other species], the rate would be affected by the new concentration of NO, resulting in a reduced reaction rate. The exact impact on the rate would depend on the order of the reaction with respect to NO.
It will be a quarter of what it was.
circumference would be halved, area would be one fourth of original , andvolume would be one eighth of original.
Kinetic energy will also be halved. Because kinetic energy is equal to 1/2 mv2
Nothing. But the current is halved.