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  • For the reaction, <img alt="\mathrm{2A_2+ B_2 \longrightarrow 2A_2B}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B2A_2&plus;%20B_2%20%5Clongrightarrow%202A_2B%7D"

For the reaction,

\mathrm{2A_2+ B_2 \longrightarrow 2A_2B}

Experimental rate law for the reaction is given as 

\mathrm{Rate = k [A_2]^2[B_2]}

What will happen to the rate of the reaction if the concentration of A2 is made 2 times while the concentration of B2 is halved?

Option: 1

Increase 4 times


Option: 2

Increase 2 times


Option: 3

Decrease 4 times


Option: 4

Decrease 2 times


Answers (1)

best_answer

Given, 

\mathrm{Rate(R)=k [A_2]^2[B_2]}

when the concentration of A2 is doubled and the concentration of B2 is halved then,

\mathrm{R'=k \left [2A_2\right ]^2 \left [ \frac{B_2}{2}\right]=2k[A]^2[B]=2R}

Hence, the new rate is 2 times the old one.

Thus, the correct answer is Option (2)

Posted by

HARSH KANKARIA

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