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How do you determine order of a reaction when the initial concentration is doubled and the initial rate is increased by a factor of 8?
Wiki User
∙ 15y ago
i think that u need to look at the rate equation which is
RATE=K*(X)*Y WHERE Y IS THE ORDER OF REACTION WRT DAT REAGENT
SO IF THE CONC INCREASED BY 2 AND UR RATE BY 8 THEN WE (2)*3=8
HENCE IT IS A THIRD ORDER RXN
IF U DONT BELIEVE ME PLOT GRAPHS OF CONC VS TIME OR REACTION RATE VS CONC
THE SHAPE TELLS US THE ORDER.....
Wiki User
∙ 15y ago
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The concentration of both A and B was doubled the reaction rate also doubled What is the rate law for this reaction?
Rate = k[A]2[B]0
When the concentration of a reactant was 0.100 M the rate of the reaction was 8.42 X 106 M s-1 When the concentration of the reactant was doubled to 0.200 M the rate increased to 1.59 X 105 M s1 What?
What indeed. Your "increased" rate has a lower value than your initial rate.
If you double a concentration of a reactant then the rate of the reaction doubled The order with respect to the reactant is?
First order; the rate is directly proportional to the concentration of reactant.
In a reversible reaction two substances are in equilibrium if the concentration of each is doubled but the equilibrium constant remains unaffected?
TRUE
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.
How is rate of a chemical reaction usually expressed?
Rates of reaction can be expressed depending upon their order.For example say you have a reaction between two chemicals and the initial rate for that reaction is known :-when:-The concentration of one of the reactants is doubled and the other reactants concentration remains the same and the overall rate of reaction does not change - reaction is zero orderwith respect to chemical which was doubled.The concentration of one of the reactants is doubled and other reactants concentration remains the same and the overall rate of reaction doubles - reaction is first order with respect to chemical which was doubled.The concentration of one of the reactants is doubled and other reactants concentration remains the same and the overall rate of reaction quadruples - reaction is second order with respect to chemical which was doubled.Zero Orderrate = kFirst Orderrate = k [A] (reaction is 1st order with respect to [A] and 1st order overall)Second Orderrate = k [A][B] (reaction is first order with respect to [A] and first order with respect to[B], reaction is second order overall)rate = k [A]2 (reaction is second order with respect to [A] and second order overall)Orders are simply added together in order to determine the overall order of reaction :-rate = k [A][B][C] would be third order overall and first order with respect to each of the reactantsThere are other orders of reaction, for example 2 and 3 quarter orders and third order reactions, but these are a little more complex.
Why is a chemical reaction rate doubled when the concentration of one reactant is doubled explain this in terms of collision theory?
Pretty simple, really. For any one "A" molecule, if there are twice as many of the other "B" molecule present then the odds of it colliding with one of them are twice as high. The same equations for effective collisions hold, so doubling the concentration doubles the reaction rate.
By how much is loudness increased if the intensity is increased 10 times?
Doubled.
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 happened to tariffs during McKinleys presidency?
the tariffs increased:]
How do you calculate turnover number for an enzyme?
As enzyme concentration increases the more active sites there are avalible, so the rate of reaction increases. therefore the turnover number increases.Hope it helped!TashaThe above it not true. The turn over number is Vmax/Et so if the enzyme concentration is doubled the velocity will also be doubled. Therefore the turn over number will remain constnat.
What is the expression for four increased a number then doubled?
2x+4
