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What would happen to the rate of a reaction with rate law rate k NO 2 H2 if the concentration of NO were halved?
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.
What would happen to the rate of a reaction with rate law rate k NO2H2 if the concentration of H2 were halved?
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.
What would happen to the rate of a reaction with rate law rate k NO 2 H2 if the concentration of H2 were halved?
Halving the concentration of H2 will decrease the rate of the reaction, assuming it is a first-order reaction with respect to H2. Since the rate law is rate = k[NO]^2[H2], cutting the concentration of H2 in half will decrease the rate of the reaction by a factor of 0.5.
What would happen to the rate of a reaction with rate law rate k NO2H2 if the concentration of NO were doubled?
The rate would be one-fourth. Correct on Apex.
What would happen to the rate of a reaction with rate law rate kno2h2 if the no were doubled?
the rate would be four times larger. apex
What would happen to the rate of a reaction with rate law rate k NO 2 H2 if the concentration of NO were halved?
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.
What would happen to the rate of a reaction with rate law rate k NO2H2 if NO were halved?
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.
What would happen to the rate of a reaction with rate law rate equals k NO 2 H2 if the concentration of NO were halved?
It will decrease by half.
What would happen to the rate of a reaction with law rate k NO2H2 if the concentration of H2 were halved?
The rate would be four times larger. Impossible.
What would happen to the rate of a reaction with rate law ratekNO2H2 if the concentratin of H2 were halved?
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.
What would happen to the rate of a reaction with rate law rate kNO2H2 if the concentration of NO were halved?
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.
What would happen to the rate of a reaction with rate law rate k NO2H2 if the concentration of H2 were halved?
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.
What would happen to the rate of a reaction with rate law rate k(NO)2(H2) if the concentration of NO were halved?
If the concentration of NO is halved in a reaction with the rate law rate = k(NO)²(H₂), the rate of the reaction would decrease. Specifically, since the rate is proportional to the square of the concentration of NO, reducing its concentration by half would result in the rate being reduced to one-fourth of its original value, assuming the concentration of H₂ remains constant. Therefore, the new rate would be k(0.5NO)²(H₂) = k(0.25NO²)(H₂) = (1/4) × original rate.
What would happen to the rate of a reaction with rate law rate k NO 2 H2 if the concentration of H2 were halved?
Halving the concentration of H2 will decrease the rate of the reaction, assuming it is a first-order reaction with respect to H2. Since the rate law is rate = k[NO]^2[H2], cutting the concentration of H2 in half will decrease the rate of the reaction by a factor of 0.5.
What is the order of reaction if the half life of a reaction is halved as the initial concentration of the reactant is doubled.?
Second order. If the half life of a reaction is halved as the initial concentration of the reactant is doubled, it means that half life is inversely proportional to initial concentration for this reaction. The only half life equation that fits this is the one for a second-order reaction. t(1/2) = 1/[Ao]k As you can see since k remains constant, if you double [Ao], you will cause t(1/2) to be halved.
What would happen to the rate of a reaction if the concentration of substrate was increased after the point of saturation?
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What do you think would happen if you halved the radius of a circumference?
circumference would be halved, area would be one fourth of original , andvolume would be one eighth of original.
