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

If the concentration of a first order reaction is increased by $x$ times,then the rate constant $(k)$ becomes

$A$ first order reaction which is $30\%$ complete in $30 \ min$ has a half-life period of .............. $min.$ (in $.2$)

$A$ first order reaction is $50 \%$ complete in $16 \ min$. How much time is required to complete $87.5 \%$ of the reaction (in $min$)?

For a certain first order reaction,$32 \%$ of the reactant is left after $570 \ s$. The rate constant of this reaction is ........... $\times 10^{-3} \ s^{-1}$. (Round off to the Nearest Integer).
$[$Given: $\log_{10} 2 = 0.301, \ln 10 = 2.303]$

The rate constant for the decomposition of $H_2O_2$ is $3.66 \times 10^{-3} \ s^{-1}$. If the initial concentration of $H_2O_2$ is $0.882 \ M$,then in how many seconds will its concentration become $0.600 \ M$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module