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Question: \(A+2B\to C\) , the rate equation for reaction is given Rate = [A][B]. if the concentration of A is ...

A+2B→CA+2B\to C , the rate equation for reaction is given Rate = [A][B]. if the concentration of A is kept constant while B is doubled. What will happen to the rate?
(A) Doubled
(B) Halved
(C) The same
(D) Quadrupled

Explanation

Solution

As the reaction proceeds, the number of reactants decreases and the number of products increase in a chemical reaction. The number of reactants consumed or the number of products formed depends on the overall rate of the reaction. The rate of formation or the rate of disappearances is equal to the overall rate of the reaction.

Complete step by step solution:
Consider a reaction,
xX+yY→aA+bBxX+yY\to aA+bB
The rate of reaction = rate of disappearances of reactant = rate of formation of products
r=−1xΔ[X]Δt=−1yΔ[Y]Δt=1aΔ[A]Δt=1bΔ[B]Δtr=-\dfrac{1}{x}\dfrac{\Delta [X]}{\Delta t}=-\dfrac{1}{y}\dfrac{\Delta [Y]}{\Delta t}=\dfrac{1}{a}\dfrac{\Delta [A]}{\Delta t}=\dfrac{1}{b}\dfrac{\Delta [B]}{\Delta t}
Where r = rate of the overall reaction
Δ[X],Δ[Y],Δ[A],Δ[B]\Delta [X],\Delta [Y],\Delta [A],\Delta [B] , are the concentrations of the reactant and products and Δt\Delta t is a change in time
Hence, therateofoverallreaction=the rate of overall reaction = rate of disappearances of reactant/rate of formation of productsstoichiometric coefficient of reactants/products\dfrac{rate\text{ }of\text{ }disappearances\text{ }of\text{ }reactant/rate\text{ }of\text{ }formation\text{ }of\text{ }products}{stoichiometric\text{ }coefficient\text{ }of\text{ }reactants/products}
Factors affecting the rate of reaction:
The following factors which alter the rate of reaction,
(i) Concentration of reactants
(ii) Temperature
(iii) Effect of solvent
(iv) Phase and surface area of reactants
The concentration of reactants: based on collision theory, reactant molecules collide with each other to form products. The number of colliding particles will increase, then the rate of reaction increases by increasing the concentration of reactants.
Given reaction, A+2B→CA+2B\to C
Given the concentration of A is kept constant and the concentration of B doubled, then the rate of reaction will be Quadrupled.
r1=[A][B]2{{r}_{1}}=[A]{{[B]}^{2}}
Concentration of A is constant, [A]=1, and concentration of B doubled, [B]= [2B]
Then, r2=[A][2B]2=4[A][B]2=4r1{{r}_{2}}=[A]{{[2B]}^{2}}=4[A]{{[B]}^{2}}=4{{r}_{1}}

The correct option is D.

Note: The number of collisions between reactant molecules per second, when the temperature is increased, thereby increases the rate of reaction. The rate of reaction by changing the reaction mechanism is a catalyst that increases or decreases the rate of reaction respectively.