Calculate the change in pressure (in atm) when 2 mole of NO and 16 gram O 2 in a 6.25 litre originally at 27° C react to produce the maximum quantity of NO 2 possible according to the equation - 2 NO ( g )+ O 2( g ) longrightarrow 2 NO 2( g ) (Take R =(1/12) L atm / mol - K )

Question

Calculate the change in pressure (in atm) when 2 mole of NO and 16 gram O 2 in a 6.25 litre originally at 27° C react to produce the maximum quantity of NO 2 possible according to the equation - 2 NO ( g )+ O 2( g ) longrightarrow 2 NO 2( g ) (Take R =(1/12) L atm / mol - K ) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/a1b3a5eda3e55190aeeeacd6e6ff7cf8-.png" /> Δ n =|2.5-2|=0.5 Change in pressure Δ P =(Δ nRT / V )=0.5 × (1/12) × (300/6.25) atm =2 atm

Q. Calculate the change in pressure (in atm) when 2 mole of NO and 16 gram O2​ in a 6.25 litre originally at 27∘C react to produce the maximum quantity of NO2​ possible according to the equation - 2NO(g)+O2​(g)⟶2NO2​(g) (Take R=121​Latm/mol−K )

Δn=∣2.5−2∣=0.5 Change in pressure ΔP=VΔnRT​=0.5×121​×6.25300​atm =2atm

$Δ n = ∣2.5 - 2∣ = 0.5$
Change in pressure
$Δ P = \frac{Δ n RT}{V} = 0.5 \times \frac{1}{12} \times \frac{300}{6.25} a t m$
$= 2 a t m$