JEE Advance - Chemistry (2019 - Paper 2 Offline - No. 9)
The decomposition reaction
$$2{N_2}{O_5}(g)\buildrel \Delta \over \longrightarrow 2{N_2}{O_4}(g) + {O_2}(g)$$
is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y $$ \times $$ 103 s, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5 $$ \times $$ 10-4s-1, assuming ideal gas behaviour, the value of Y is ...............
$$2{N_2}{O_5}(g)\buildrel \Delta \over \longrightarrow 2{N_2}{O_4}(g) + {O_2}(g)$$
is started in a closed cylinder under isothermal isochoric condition at an initial pressure of 1 atm. After Y $$ \times $$ 103 s, the pressure inside the cylinder is found to be 1.45 atm. If the rate constant of the reaction is 5 $$ \times $$ 10-4s-1, assuming ideal gas behaviour, the value of Y is ...............
Answer
2.3
Explanation
At constant V, T
At initial t = 0 and final t = y $$ \times $$ 103 sec
$$\matrix{ {2{N_2}{O_5}(g)\buildrel \Delta \over \longrightarrow } & {2{N_2}{O_4}(g) + } & {{O_2}(g)} \cr 1 & 0 & 0 \cr {1 - 2p} & {2p} & p \cr } $$
pTotal = 1 $$-$$ 2p + 2p + p
1.4 = 1 + p
p = 0.45 atm
According to first order reaction,
$$k = {{2.303} \over t}\log {{{p_i}} \over {{p_i} - 2p}}$$
pi = 1 atm (given)
2p = 2 $$ \times $$ 0.45 = 0.9 atm
On substituting the values in above equation,
$$2k.t = 2.303\log {1 \over {1 - 0.9}}$$
$$2 \times 5 \times {10^{ - 4}} \times y \times {10^3} = 2.303\log {1 \over {0.1}}$$
$$y = 2.303 = 2.3$$
Note : Unit of rate constant (k), i.e. s$$-$$1 represents that it is a first order reaction.
At initial t = 0 and final t = y $$ \times $$ 103 sec
$$\matrix{ {2{N_2}{O_5}(g)\buildrel \Delta \over \longrightarrow } & {2{N_2}{O_4}(g) + } & {{O_2}(g)} \cr 1 & 0 & 0 \cr {1 - 2p} & {2p} & p \cr } $$
pTotal = 1 $$-$$ 2p + 2p + p
1.4 = 1 + p
p = 0.45 atm
According to first order reaction,
$$k = {{2.303} \over t}\log {{{p_i}} \over {{p_i} - 2p}}$$
pi = 1 atm (given)
2p = 2 $$ \times $$ 0.45 = 0.9 atm
On substituting the values in above equation,
$$2k.t = 2.303\log {1 \over {1 - 0.9}}$$
$$2 \times 5 \times {10^{ - 4}} \times y \times {10^3} = 2.303\log {1 \over {0.1}}$$
$$y = 2.303 = 2.3$$
Note : Unit of rate constant (k), i.e. s$$-$$1 represents that it is a first order reaction.
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