The pressure of iodine gas at a particular temperature is found to be `0.111atm`, where as the expected pressure is `0.074 atm`, the increased pressure is due to `I_(2)hArr 2I`. Calculate `K_(p)` for this equilibrium.
The pressure of iodine gas at 1273 K is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation I_(2) hArr 2I . Calculate K_(p) .
At 1000^@C the pressure of iodine gas is found to be 0.112 atm whereas the expected pressure is 0.074 atm. The increased pressure is due to dissociation, I_2 hArr 2I . Calculate K_p . Also find out pressure at which I_2 will be 90% dissociation at 1000^@C .
At a certain temperature and a total pressure of 10^(5) Pa, iodine vaour contain 40% by volume of iodine atmos [I_(2) (g) hArr 2 I (g) ]. " Calculate " K_(p) for the equilibrium.
At 1000K, the pressure of iodine gas is found to be 0.112 atm due to partial dissociation of I_(2)(g) into I(g). Had there been no dissociation, the pressure would have been 0.074 atm. Calculate the value of K_(p) for the reaction: I_(2)(g) hArr 2I(g) .
At a certain temperature and a total pressure of 10^(5) Pa , iodine vapour contains 40% by volume of I "atoms" , Calculate K_(p) for the equilibrium. I_(2(g))hArr2I_((g))
Osmotic pressure of insulin solution at 298 K is found to be 0.0072atm . Hence, height of water Column due to this pressure is
At 1000K , the pressure of iodine gas is found to be 0.1 atm due to partial dissociation of I_(2)(g) into I(g) . Had there been no dissociation, the pressure would have been 0.07 atm . Calculate the value of K_(p) for the reaction: I_(2)(g)hArr2I(g) .