`K_(p)` for equilibrium `N_(2)O_(4) hArr 2NO_(2)` is `0.25` at `15^(@)C`. If the system is allowed to expand `&` `N_(2)` is added at a constant pressure of 1 atm. What will be the degree when partial of `N_(2)` is 0.6 atm.
A
`0.38`
B
`0.23`
C
`0.61`
D
`0.55`
Text Solution
Verified by Experts
The correct Answer is:
1
`{:(N_(2)O_(4)hArr,2NO_(2),,P+P=0.4atm,,P_(N_(2))=0.6atm),(P_(0),-,,K_(p)=(P^(2))/(P)=0.25,,),(P_(0)-P'//2=P,P,,P=4P^(2),,),(4P'^(2)+P'-0.4=0,,,P=0.18,,),(P'=0.22,,,,,):}` Degree of dissociation `=(P'//2)/(P_(0))=(0.11)/(0.18+0.11)=0.38`
2.8 g of N_(2) gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm. Calculate W for the gas.
K_(p) for the reaction N_(2) O_(4) (g) rarr 2NO_(2)(g) is 0.66 at 46^(@)C . Calculate the percent dissociation of N_(2)O_(4) at 46^(@)C and a total pressue of 0.5 atm. Also calculate the partial pressure of N_(2)O_(4) and NO_(2) at equilibrium.
28g of N_(2) gas at 300 K and 20 atm was allowed to expand isothermally against a constant external pressure of 1 atm , q for the gas is (R=0.082) .
2.8g of N_(2) gas at 300K and 20atm was allowed to expand isothermally against a constant external pressure of 1atm . Calculate DeltaU, q, and W for the gas.
K_p for the reaction : N_2O_4(g) hArr 2NO_2(g) is 0.157 atm at 27^@C and 1 atm pressure . Calculate K_c for the reaction.
For the following equilibrium N_(2)O_(4)(g)hArr 2NO_(2)(g) K_(p) is found to be equal to K_(c) . This is attained when :
For the dissociation reaction N_(2)OhArr2NO_(2)(g), the equilibrium constant K_(P) is 0.120 atm at 298 K and total pressure of system is 2 atm. Calculate the degree of dissociation of N_(2)O_(4) .
In a container equilibrium N_(2)O_(4) (g)hArr2NO_(2) (g) is attained at 25^(@)C . The total equilibrium pressure in container is 380 torr. If equilibrium constant of above equilibrium is 0.667 atm, then degree of dissociation of N_(2)O_(4) at this temperature will be:
