JEE Main 2014 · 19 Apr · Shift 1 · Chemistry Q36

ChemistrySingle correct+4 / -1Question 36 of 90

For the reaction,
$\ce{3A + 2B -> C + D},$
the differential rate law can be written as :

\dfrac13\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k[A]^n[B]^m

-\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k[A]^n[B]^m

+\dfrac13\dfrac{d[A]}{dt}=-\dfrac{d[C]}{dt}=k[A]^n[B]^m

-\dfrac13\dfrac{d[A]}{dt}=\dfrac{d[C]}{dt}=k[A]^n[B]^m

Answer and worked solution

Official answer: D

Rate $=-\dfrac13\dfrac{d[A]}{dt}=-\dfrac12\dfrac{d[B]}{dt}=\dfrac{d[C]}{dt}=\dfrac{d[D]}{dt}=k[A]^n[B]^m$ . This is option (4).

JEE Main 2014 · 19 Apr · Shift 1, Chemistry question 36