In a hydrogenation reaction at 27,^oC,it is o | vedclass

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(B) The rate of reaction in terms of pressure is given by $\frac{dp}{dt} = \frac{P_1 - P_2}{\Delta t} = \frac{2 - 1.2}{50 	imes 60} \, atm \, s^{-1}

Vedclass · Accepted Answer

$1.08 	imes 10^{-5} \, M s^{-1}$

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(B) The rate of reaction in terms of pressure is given by $\frac{dp}{dt} = \frac{P_1 - P_2}{\Delta t} = \frac{2 - 1.2}{50 	imes 60} \, atm \, s^{-1} = \frac{0.8}{3000} \, atm \, s^{-1} = 2.66 	imes 10^{-4} \, atm \, s^{-1}$.Using the ideal gas equation $PV = nRT$,we have $P = CRT$,so $\frac{dP}{dt} = RT \frac{dC}{dt}$.Here,$R = 0.0821 \, L \, atm \, K^{-1} \, mol^{-1}$ and $T = 27 + 273 = 300 \, K$.Therefore,$\frac{dC}{dt} = \frac{1}{RT} 	imes \frac{dP}{dt} = \frac{2.66 	imes 10^{-4}}{0.0821 	imes 300} \, M s^{-1}$.$\frac{dC}{dt} = \frac{2.66 	imes 10^{-4}}{24.63} \approx 1.08 	imes 10^{-5} \, M s^{-1}$.

In a hydrogenation reaction at $27\,^oC$,it is observed that the hydrogen gas pressure falls from $2\,atm$ to $1.2\,atm$ in $50\,min$. Calculate the rate of disappearance of hydrogen.

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