At `60^(@)" "N_(2)O_(4)` is `50%` dissociated at this temperature and one atmosphere pressure the standard free energy change is

At 60^(@) and 1 atm, N_(2)O_(4) is 50% dissociated into NO_(2) then K_(p) is

At 60^(@) and 1 atm, N_(2)O_(4) is 50% dissociated into NO_(2) then K_(p) is

N_(2) O_(4) is 25% dissociated at 37^(@)C and one atmosphere pressure. Calculate (i) K_(p) and (ii) the percentage dissociation at 0.1 atm and 37^(@)C

At 60^(@_C and 1 atm, N_2O_4 is 50% dissociated into NO_2 then K_p is -

N_(2)_(4) is 25% dissociated at 37^(@)C and one atmosphere pressure. Calculate (i) K_(p) and (ii) the percentage dissociation at 0.1 atm and 37^(@)C

PCl_(5) is 50% dissociated into PCl_(3) and Cl_(2) at 1 atmosphere pressure. It will be 40% dissociated at:

A reaction is spontaneous if free energy change Delta G , for the reaction is negative . The question arises, how we can explain , the spontaneous reversible reaction taking the example of dissociation equilibrium , N_2O_4 hArr 2NO_2(g) . the variation of free energy with the fraction of N_2O_4 dissociated under standard conditions is shown in the figure below. The standard free energy change for the conversion of 1 mole of N_2O_4 into 2 moles of NO_2 is