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What would happen to the rate of a reaction with rate law rate k (NO2) (H2) if the concentration of Anna were doubled?
The rate of the reaction with the rate law ( \text{rate} = k[\text{NO}_2][\text{H}_2] ) is dependent solely on the concentrations of NO2 and H2. If the concentration of Anna, which is not included in the rate law, is doubled, it would have no effect on the rate of the reaction. The reaction rate would remain unchanged unless the rate law itself involves Anna in some way, which is not indicated here.
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What would happen to the rate of a reaction with rate law rate kno2h3 if the concentration of no were doubled?
If the concentration of NO was doubled in the rate law rate = k[NO]2[H3], the rate of the reaction would increase by a factor of 4. This is because the rate of a reaction typically increases with an increase in the concentration of reactants, raised to a power dictated by their respective coefficients in the rate law equation.
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 with rate law rate k NO H2 if the concentrations of NO were doubled?
The rate would be four times larger
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 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 kno2h3 if the concentration of no were doubled?
If the concentration of NO was doubled in the rate law rate = k[NO]2[H3], the rate of the reaction would increase by a factor of 4. This is because the rate of a reaction typically increases with an increase in the concentration of reactants, raised to a power dictated by their respective coefficients in the rate law equation.
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 kNO2H2 if the concentration of NO were doubled?
The rate would be four times larger
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 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 if the concentration of substrate was increased after the point of saturation?
zff
What would happen to the rate of a reaction with rate law rate k NO H2 if the concentrations of NO were doubled?
The rate would be four times larger
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 k no 2 h2 if the concentration of h2 were halved-?
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.
What would happen to the rate of a reaction with rate law rate kNO2H2 if the concentration of NO were double?
The rate would quadruple (increase by a factor of 4). This is because the rate depends on the SQUARE of the concentration of NO.
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.
