The gaseous reaction A_{(g)} to 2B_{(g)} + C_ | vedclass

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

Question

(A) For the reaction $A_{(g)} 	o 2B_{(g)} + C_{(g)}$,let the initial pressure of $A$ be $P_0 = 90 \ mm \ Hg$.At $t = 10 \ min$,let the decrease in pr

Vedclass · Accepted Answer

$1.15 	imes 10^{-3} \ s^{-1}$

Vedclass · Answer

(A) For the reaction $A_{(g)} 	o 2B_{(g)} + C_{(g)}$,let the initial pressure of $A$ be $P_0 = 90 \ mm \ Hg$.At $t = 10 \ min$,let the decrease in pressure of $A$ be $x$.Then,the partial pressures are: $P_A = P_0 - x$,$P_B = 2x$,and $P_C = x$.The total pressure $P_t = P_A + P_B + P_C = (P_0 - x) + 2x + x = P_0 + 2x$.Given $P_t = 180 \ mm \ Hg$ and $P_0 = 90 \ mm \ Hg$,we have $180 = 90 + 2x$,which gives $2x = 90$,so $x = 45 \ mm \ Hg$.The pressure of $A$ at time $t$ is $P_A = 90 - 45 = 45 \ mm \ Hg$.The rate constant $k$ for a first-order reaction is given by $k = \frac{2.303}{t} \log \left( \frac{P_0}{P_A} ight)$.Converting time to seconds: $t = 10 \ min = 600 \ s$.$k = \frac{2.303}{600} \log \left( \frac{90}{45} ight) = \frac{2.303}{600} \log(2) = \frac{2.303 	imes 0.3010}{600} \approx 1.155 	imes 10^{-3} \ s^{-1}$.

The gaseous reaction $A_{(g)} \to 2B_{(g)} + C_{(g)}$ is a first-order reaction. If the initial pressure $P_A = 90 \ mm \ Hg$ and the total pressure after $10 \ min$ is $180 \ mm \ Hg$,calculate the rate constant of the reaction.

Similar Questions

For the first$-order$ reaction,$T_{av}$ (average life),$T_{50}$ and $T_{75}$ in the increasing order are:

For the first order reaction,half life is $14 \, sec$. The time required for the initial concentration to reduce to $\frac{1}{8}$ of its value is .......... $sec$.

For a first-order reaction,the rate constant is $4 \times 10^{-3} \, s^{-1}$. If the concentration of the reactant is $0.02 \, M$,what will be the rate of the reaction?

For a first order reaction,$A \to P$,$t_{1/2}$ (half-life) is $10 \ days$. The time required for $\frac{1}{4}$ conversion of $A$ (in days) is: $(\ln 2 = 0.693, \ln 3 = 1.1)$.

The half-life of a first order reaction $X \to Y$ is $100 \ min$. The concentration of $X$ would be reduced to $10 \%$ of the initial concentration in .......... $min$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module