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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 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.
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What would happen to the rate of a reaction with rate law rate kNO2H2 if the concentration of H2 were halved?
In the rate law given as rate = k[NO2][H2], the rate of the reaction is directly proportional to the concentration of H2. If the concentration of H2 is halved, the rate of reaction would also be halved, assuming the concentration of NO2 remains constant. Thus, the overall reaction rate would decrease to 50% of its original value.
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 kNO2H2 if the concentration of H2 were halved?
In the rate law ( \text{Rate} = k[\text{NO}_2][\text{H}_2] ), the rate of the reaction is directly proportional to the concentration of both reactants. If the concentration of ( \text{H}_2 ) is halved, the rate of the reaction would also decrease by half, assuming the concentration of ( \text{NO}_2 ) remains constant. This is because the rate depends on the product of the concentrations of the reactants, so any reduction in ( [\text{H}_2] ) directly affects the overall reaction rate.
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 rate law rate kNO2H2 if the concentration of H2 were halved?
In the rate law given as rate = k[NO2][H2], the rate of the reaction is directly proportional to the concentration of H2. If the concentration of H2 is halved, the rate of reaction would also be halved, assuming the concentration of NO2 remains constant. Thus, the overall reaction rate would decrease to 50% of its original value.
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 kNO2H2 if the concentration of H2 were halved?
In the rate law ( \text{Rate} = k[\text{NO}_2][\text{H}_2] ), the rate of the reaction is directly proportional to the concentration of both reactants. If the concentration of ( \text{H}_2 ) is halved, the rate of the reaction would also decrease by half, assuming the concentration of ( \text{NO}_2 ) remains constant. This is because the rate depends on the product of the concentrations of the reactants, so any reduction in ( [\text{H}_2] ) directly affects the overall reaction rate.
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 rate law kNO2H2 if the concentration of NO were doubled?
The rate would be four times larger
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
