In a second order reaction, when the concentration of both the reactants are equal, the reaction is completed `20%` in 500 s. How long would it take f

Correct Answer - A 75 % completed in 32 minutes I .e., 2 half - life has passed `therefore ` 1 half - life = 16 minutes

Correct Answer - C c) The data suggests that reaction rate ( r) `propto [A]^(2)` and is `propto [B]`. `therefore r=k[A]6(2)[B]`

Correct Answer - C c) rate = `k[CO]^(2)` By doubling the concentration of carbon monoxide (CO),the rate of reaction will become four times.

Correct Answer - B c) `t_(1//2) propto 1/([A_(0)]^(n-1)),(t_(1//2))_(1)/(t_(1//2))_(2) = [A_(0)]_(2)^(n-1)/([A_(0)]_(1)^(n-1))={([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)` `t/(t_(1//2))=(2a)/(a)^(n-1)` or `2=(2)^(n-1)` or n-1 or n=2

Correct Answer - D The time taken for half the reaction to complete i.e., the time in which the concentration of a reactant is reduced to half of its origianal value...

Correct Answer - B `xto` product (zero order) Till the concentration of X changes from `X_0` to `X_0/2` , it is zero order with rate constant k. Time taken for concentration...

The second rorder equation when both the recants have same concentration is `k=(1)/(t)*(x)/(a(a-x))` if a=100, x=20 t=500 seconds. So, `k=(1)/(500)xx(20)/(100(100-20))` When `a=100, x=60 ,t=?` `t=(1)/(lambda)*(60)/(100xx40)` Substituting the value of K,...

Correct Answer - C `k=(2.030)/(t)log.((a)/(a-x))` `k=(2.303)/(1) log. ((100)/(5))=2xx0.693` `t_(1//2)=(0.693)/(k)=(0.693)/(2xx0.693)=(1)/(2)hr`

Correct Answer - C Statement (1) cannot be correct because the rate of zero order reactions does not depend on concentrartion of reactant

Correct Answer - D Consider a hypothetical change. `A+BhArrC+D` For the reaction. `K_(eq)=([C][D])/([A][B])` For the above reaction if concentration of reactions are doubled then the rate of forward reaction increases for...