You visited us 7 times! Enjoying our articles? Unlock Full Access!

First Order Reaction

For a first o...

Question

For a first order gas phase reaction:

$A_{(g)} \to 2 B_{(g)} + C_{(g)}$ .

$P_{0}$ be initial pressure of $A$ and $P$ , the total pressure at time ' $t^{'}$ . Integrated rate equation is:

\frac{2.303}{t} log (\frac{P_{0}}{P_{0} - P_{t}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2.303}{t} log (\frac{2 P_{0}}{3 P_{0} - P_{t}})

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{2.303}{t} log (\frac{P_{0}}{2 P_{0} - P_{t}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2.303}{t} log (\frac{2 P_{0}}{2 P_{0} - P_{t}})

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{2.303}{t} log (\frac{2 P_{0}}{3 P_{0} - P_{t}})$
The given reaction is :

$A_{(g)} \to 2 B_{(g)} + C_{(g)}$

$A$ $B$ $C$
Initial $P_{0}$ $0$ $0$
Final $P_{0} - P$
$2 P$ $P$
Total pressure at time $(t) = P_{0} - P + 2 P + P = P_{t}$

$\Rightarrow P_{t} = P_{0} + 2 P$

$P_{t} - P_{0} = 2 P \Rightarrow P = \frac{P_{t} - P_{0}}{2}$

$k = \frac{2.303}{t} log [\frac{P_{0}}{P_{0} - P}] = \frac{2.303}{t} log ⎡ ⎢ ⎢ ⎢ ⎢ ⎣ \frac{P_{0}}{P_{0} - (\frac{P_{t} - P_{0}}{2})} ⎤ ⎥ ⎥ ⎥ ⎥ ⎦$

$= \frac{2.303}{t} log (\frac{2 P_{0}}{2 P_{0} - P_{t} + P_{0}}) = \frac{2.303}{t} log (\frac{2 P_{0}}{3 P_{0} - P_{t}})$

Hence, option $B$ is correct.

Suggest Corrections

Similar questions

If [A] is the concentration of reactant A at any time t and [A_o] is the concentration at t=0, then for the first order reaction the rate equation can be written as:

Q. For a first order reaction,

A (g) \to 2 B (g) + C (g)

, the rate constant in terms of initial pressure

P_{0}

and pressure at time

t (P_{t})

, is given by:

Q. A polyatomic gas

(γ = \frac{4}{3})

is compressed to

\frac{1}{8}

of its volume adiabatically and reversibly. If its initial pressure is

P_{0}

its new pressure will be

Q. Figure shows a large closed cylindrical tank containing water. Initially, the air trapped above the water surface has a height

h_{0}

and pressure

2 P_{0}

where

P_{0}

is the atmospheric pressure. There is a hole in the wall of the tank at a depth

h_{1}

below the top from which water comes out. A long vertical tube is connected as shown.

Find the height

h_{2}

of the water in the long tube above the top initially.

Q. One mole of an ideal gas undergoes a process

P = \frac{P_{0}}{1 + {(\frac{V_{0}}{V})}^{2}}

Here

P_{0}

and

V_{0}

are constants. Change in temperature of the gas when volume is changed from

V = V_{0}

V = 2 V_{0}

is :-

Join BYJU'S Learning Program

For a first order gas phase reaction: A(g)→2B(g)+C(g). P0 be initial pressure of A and P, the total pressure at time 't′. Integrated rate equation is:

For a first order gas phase reaction:

$A_{(g)} \to 2 B_{(g)} + C_{(g)}$ .

$P_{0}$ be initial pressure of $A$ and $P$ , the total pressure at time ' $t^{'}$ . Integrated rate equation is: