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When you decrease the concentration of reactants in a chemical reaction, the reaction rate typically slows down because there are fewer reactant particles available to collide and react with each other. This aligns with the principles of collision theory, which states that reactions occur when particles collide with sufficient energy and proper orientation. Consequently, the equilibrium position may also shift to favor the formation of reactants, according to Le Chatelier's principle, if the reaction is reversible.
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In a first-order reaction how does the rate change if the concentration of the reactant decreases to one-third its original value?
In a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. If the concentration decreases to one-third of its original value, the rate of the reaction will also decrease to one-third. This is because the rate equation for a first-order reaction can be expressed as ( \text{Rate} = k[A] ), where ( k ) is the rate constant and ([A]) is the concentration of the reactant. Therefore, a decrease in concentration leads to a proportional decrease in the reaction rate.
What is the decrease in the concentration of reactants over time known as?
The decrease in the concentration of reactants over time is known as the reaction rate. This term refers to the speed at which reactants are converted into products in a chemical reaction. The reaction rate can be influenced by various factors, including temperature, concentration, and the presence of catalysts.
Why first order reaction never completes?
A first-order reaction will never be completed because the reaction rate depends only on the concentration of one reactant. As the reaction progresses and the reactant is consumed, the concentration of the reactant decreases, causing the reaction rate to also decrease. This gradual decrease in reaction rate means that the reaction will continue indefinitely, given enough time.
What would happen to the rate of a reaction with rate law rate kNO2 H2 if the concentration of no were halved?
In the given rate law, the rate of the reaction is proportional to the concentration of NO2 raised to the power of 1 and the concentration of H2 raised to the power of 1, represented as rate = k[NO2][H2]. If the concentration of NO2 is halved, the rate of the reaction would also be halved, assuming the concentration of H2 remains constant. This is because the rate directly depends on the concentration of NO2, leading to a linear decrease in the reaction rate with a decrease in its concentration.
How in terms of collision theory how a decrease in concentration can change the rate of a precipitation reaction?
Decrease in conc. means decrease in no. of molecules per unit volume. Since, molecules or atoms will less in number, the no. of colliding particles will be less. This will lead to decreased precipitation
What causes the reaction rate to decrease as the reaction progress?
The concentration of the reactants decreases.
In a first-order reaction how does the rate change if the concentration of the reactant decreases to one-third its original value?
In a first-order reaction, the rate of reaction is directly proportional to the concentration of the reactant. If the concentration decreases to one-third of its original value, the rate of the reaction will also decrease to one-third. This is because the rate equation for a first-order reaction can be expressed as ( \text{Rate} = k[A] ), where ( k ) is the rate constant and ([A]) is the concentration of the reactant. Therefore, a decrease in concentration leads to a proportional decrease in the reaction rate.
What is the decrease in the concentration of reactants over time known as?
The decrease in the concentration of reactants over time is known as the reaction rate. This term refers to the speed at which reactants are converted into products in a chemical reaction. The reaction rate can be influenced by various factors, including temperature, concentration, and the presence of catalysts.
Why does the reaction decrease as the reaction progresses?
There are fewer reactants left to collide.
Why first order reaction never completes?
A first-order reaction will never be completed because the reaction rate depends only on the concentration of one reactant. As the reaction progresses and the reactant is consumed, the concentration of the reactant decreases, causing the reaction rate to also decrease. This gradual decrease in reaction rate means that the reaction will continue indefinitely, given enough time.
What would happen to the rate of a reaction with rate law rate kNO2 H2 if the concentration of no were halved?
In the given rate law, the rate of the reaction is proportional to the concentration of NO2 raised to the power of 1 and the concentration of H2 raised to the power of 1, represented as rate = k[NO2][H2]. If the concentration of NO2 is halved, the rate of the reaction would also be halved, assuming the concentration of H2 remains constant. This is because the rate directly depends on the concentration of NO2, leading to a linear decrease in the reaction rate with a decrease in its concentration.
What causes the reaction rate to decrease as the reaction progresses?
The concentration of the reactants decreases.
How in terms of collision theory how a decrease in concentration can change the rate of a precipitation reaction?
Decrease in conc. means decrease in no. of molecules per unit volume. Since, molecules or atoms will less in number, the no. of colliding particles will be less. This will lead to decreased precipitation
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 ratek NO2 H2 if the concentration of H2 were halved?
In the rate law given as rate = k[NO2][H2], the reaction rate is directly proportional to the concentration of both NO2 and H2. If the concentration of H2 is halved, the reaction rate would also be halved, assuming the concentration of NO2 remains constant. This is because the rate depends linearly on the concentration of H2, so any decrease in H2 concentration results in a proportional decrease in the overall reaction 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 effects does the concentration of reactants have on the rate of the reaction?
increasing the concentration increases the rate of the reaction
