Calculate the change in pressure ( in atm) when 2 mole of `NO` and `16 g O _(2)` in `6.25` litre originally at `27^(@)C` react to produce the maximum quantity of `NO_(2)` possible according to the equation. `( ` Take `R=(1)/(12)` ltr. Atm`//`mol K `)` `2NO(g)+O_(2)(g) hArr 2NO_(2)(g)`
A
1
B
4
C
5
D
2
Text Solution
Verified by Experts
The correct Answer is:
4
`{:(,2NO,+,o_(2)(g),rarr,2NO_(2)(g)),("moles",2,,0.5,,0),(,2-2xx0.5,,0,,2xx0.5),(,1,,0,,1),(,n_(f)=1+1,,,,n_(f)=2+0.5),( :. ,Deltan=(2.5-2),=0.5"moles",,,),( :. ,"change in pressure",,,,):} ` `Deltap=(DeltanRT)/(V)=0.5xx(1)/(12)xx(3300)/(8.25)=2atm`
Calculate the change in pressure when 1.04 mole of NO and 20.0gO_(2) in a 20.0L vessel originally at 27^(@)C react to produce maximum quantity of NO_(2(g)) according to reaction 2NO_((g)) +O_(2(g))rarr2NO_(2(g)) .
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2 moles of NO(g) and 16 gm of O_(2)(g) were mixed in a 6.25 litre vessel at 27^(@)C temperature to produce maximum amount of NO_(2)(g) 2NO(g)+O_(2)(g)rarr2NO_(2)(g) ("Use" :R=(1)/(12) "L-atm mol"^(-1)K^(-1)) Calculate change in pressure (in atm) due to this reaction.
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N_(2)O_(3) dissociates into NO and NO_(2) . At equilibrium pressure of 3 atm , all three gases were found to have equal number of moles in a vessel. In another vessel, equimolormixture of N_(2)O_(3), NO and NO_(2) are taken at the same temperature but at an initial pressure of 9 atm then find the partial pressure of NO_(2) (in atm ) at equilibrium in second vessel! N_(2)O_(3)(g) hArr NO(g) + NO_(2) (g)