What happens to rate of reaction if one of the reactant is doubled
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here is ur answer✍️✍️✍️
We know, Rate of reaction is directly proportional to the concentration of the reactant raised to its order.
=>Rate=k∗[A]n=>Rate=k∗[A]n
where k is the rate constant, [A] is the concentration and n is the order of the reaction.
Given, New rate = 2 * Old rate, new concentration = 4 * old concentration.
=>k∗(4∗[A])n=2∗k∗[A]n=>k∗(4∗[A])n=2∗k∗[A]n
=>4n=2=>4n=2
.: n = 1/21/2
Therefore the order is 0.5
hope it helps u❤️❤️
We know, Rate of reaction is directly proportional to the concentration of the reactant raised to its order.
=>Rate=k∗[A]n=>Rate=k∗[A]n
where k is the rate constant, [A] is the concentration and n is the order of the reaction.
Given, New rate = 2 * Old rate, new concentration = 4 * old concentration.
=>k∗(4∗[A])n=2∗k∗[A]n=>k∗(4∗[A])n=2∗k∗[A]n
=>4n=2=>4n=2
.: n = 1/21/2
Therefore the order is 0.5
hope it helps u❤️❤️
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