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 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 .

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 20L of a gas kept at 300 K is found to be 2 atm. If the pressure is in-creased to 3 atm at the same temperauture, What wil be the new volume?

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 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) .

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) .