Azo-isopropane decomposes according to the fo | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the reaction: $((CH_3)_2CHN = NCH(CH_3)_2)_{(g)} \rightarrow (N_2)_{(g)} + (C_6H_{14})_{(g)}$At $t=0$: Initial pressure = $P_o$,$N_2$ = $0$,$C

Vedclass · Accepted Answer

$K = \frac{2.303}{t} \log \frac{P_o}{2P_o - P_t}$

Vedclass · Answer

(A) For the reaction: $((CH_3)_2CHN = NCH(CH_3)_2)_{(g)} \rightarrow (N_2)_{(g)} + (C_6H_{14})_{(g)}$At $t=0$: Initial pressure = $P_o$,$N_2$ = $0$,$C_6H_{14}$ = $0$At time $t$: Pressure = $(P_o - x)$,$N_2$ = $x$,$C_6H_{14}$ = $x$The total pressure $P_t$ at time $t$ is given by:$P_t = (P_o - x) + x + x = P_o + x$Therefore,$x = P_t - P_o$The partial pressure of the reactant at time $t$ is $(P_o - x) = P_o - (P_t - P_o) = 2P_o - P_t$For a first-order reaction,the rate constant $K$ is:$K = \frac{2.303}{t} \log \frac{P_o}{P_o - x}$Substituting $(P_o - x) = 2P_o - P_t$:$K = \frac{2.303}{t} \log \frac{P_o}{2P_o - P_t}$

Similar Questions

For a first order reaction $A \to B$,the reaction rate at reactant concentration of $0.01 \ M$ is found to be $2.0 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$. The half-life period of the reaction is .......... $sec$.

For a first-order reaction,the rate constant is $60 \ s^{-1}$. How much time will it take for the initial concentration of the reactant to be reduced to $1/10$th of its initial value?

For a reaction $A \rightarrow$ product,the rate constant is $2 \times 10^{-2} \ s^{-1}$. The initial concentration of $A$ is $1.0 \ mol \ dm^{-3}$. What is the value of $\log \frac{1}{[A]_{t}}$ after $100 \ s$?

After how many seconds will the concentration of the reactants in a first order reaction be halved,if the decay constant is $1.155 \times 10^{-3} \, \sec^{-1}$?

In a first order reaction,the concentration of the reactant is reduced to $25\%$ in one hour. The half-life period of the reaction is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module