Solveeit Logo
Login

Question

Chemistry Question on Chemical Kinetics

For the reaction, 3A + 2B →\to C + D, the differential rate law can be written as :

A

13d[A]dt=d[C]dt=k[A]n[B]m\frac{1}{3} \frac{d\left[A\right]}{dt} = \frac{d\left[C\right]}{dt} = k\left[A\right]^{n} \left[B\right]^{m}

B

−d[A]dt=d[C]dt=k[A]n[B]m- \frac{d\left[A\right]}{dt} = \frac{d\left[C\right]}{dt} = k\left[A\right]^{n} \left[B\right]^{m}

C

+13d[A]dt=d[C]dt=k[A]n[B]m+\frac{1}{3} \frac{d\left[A\right]}{dt} = \frac{d\left[C\right]}{dt} = k\left[A\right]^{n} \left[B\right]^{m}

D

−13d[A]dt=d[C]dt=k[A]n[B]m-\frac{1}{3} \frac{d\left[A\right]}{dt} = \frac{d\left[C\right]}{dt} = k\left[A\right]^{n} \left[B\right]^{m}

Answer

−13d[A]dt=d[C]dt=k[A]n[B]m-\frac{1}{3} \frac{d\left[A\right]}{dt} = \frac{d\left[C\right]}{dt} = k\left[A\right]^{n} \left[B\right]^{m}

Explanation

Solution

Rate =−13d(A)dt=−12d(B)dt=d(C)dt=d(D)dt=-\frac{1}{3} \frac{ d ( A )}{ dt }=-\frac{1}{2} \frac{ d ( B )}{ dt }=\frac{ d ( C )}{ dt }=\frac{ d ( D )}{ dt }

Rate =K(A)n(B)n= K ( A )^{ n }( B )^{ n }
−13d(A)dt=−12d(B)dt=d(C)dt=d(D)dt-\frac{1}{3} \frac{ d ( A )}{ dt } =-\frac{1}{2} \frac{ d ( B )}{ dt }=\frac{ d ( C )}{ dt }=\frac{ d ( D )}{ dt }
=K(A)n(B)m= K ( A )^{ n }( B )^{ m }

so −13d(A)dt=d(C)dt-\frac{1}{3} \frac{ d ( A )}{ dt }=\frac{ d ( C )}{ dt }
=K(A)n(B)m= K ( A )^{ n }( B )^{ m }