Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

For the reaction 2A+B \rightarrow C, the values of initial rate at different reactant concentrations are given in the table below. The rate law for the reaction is :

[A](mol\; L^{-1}) [B](mol\; L^{-1}) \\Initial Rate\\(mol\; L^{-1}s^{-1})
0.05 0.05 0.045
0.10 0.05 0.090
0.20 0.10 0.72

 

  • Option 1)

    Rate=k[A]^{2}[B]^{2}

  • Option 2)

    Rate=k[A]^{2}[B]

  • Option 3)

    Rate=k[A][B]^{2}

     

  • Option 4)

    Rate=k[A][B]

 

Answers (1)

best_answer

Rate law\Rightarrow R=k[A]^{x}[B]^{y}

[A](mol\; L^{-1}) [B](mol\; L^{-1}) \\Initial Rate\\(mol\; L^{-1}s^{-1})
0.05 0.05 0.045 R_{1}
0.10 0.05 0.090 R_{2}
0.20 0.10 0.72 R_{3}

R_{1}=0.045=k[0.05]^{x}[0.05]^{y}

R_{2}=0.09=k[0.01]^{x}[0.05]^{y}

\frac{R_{2}}{R_{1}}=2=2^{x}\; \; \; \; x=1               \frac{R_{3}}{R_{2}}=\frac{0.72}{0.09}=\frac{k[0.2]^{x}}{k[0.1]^{x}}\frac{k[0.1]^{x}}{k[0.05]^{x}}

y=2

\therefore Rate \; law:-R=k[A][B]^{2}


Option 1)

Rate=k[A]^{2}[B]^{2}

Option 2)

Rate=k[A]^{2}[B]

Option 3)

Rate=k[A][B]^{2}

 

Option 4)

Rate=k[A][B]

Posted by

solutionqc

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: NIT Rourkela records higher closing ranks in CSE, ECE, AI; core engineering branches stable
anam-khan

JEE Main 2026: IIT Delhi’s CSE, EE most opted; last 2 year’s closing ranks for all categories
anam-khan

Must provide CBSE 11th registration numbers to JEE Mains aspirants, board tells schools

Latest Question