The hydrogenation of vegetable ghee at `25^(@)C` reduces the pressure of `H_(2)` form `2 atm` to `1.2 atm` in `50min`. Calculate the rate of reaction in terms of change of
(a) Pressure per minute
(b) Molarity per second

In hydrogenation reaction at 25^(@)C , it is observed that hydrogen gas pressure falls from 2 atm to 1.2 atm isn 50 min. Calculate the rate of reaction in molarity per sec. (R=0.0821 litre -atm "degree"^(-1) "mol"^(-1))

At 27^(@)C it was observed in the hydrogenation of a reaction, the pressure of H_(2)(g) decreases form 10 atm to 2 atm in 10 min . Calculate the rate of reaction in M min^(-1) (Given R = 0.08 L atm K^(-1) mol^(-1) )

For the decompoistion of phosphorus pentachloride at 200^(@)C , it is observed that its pressure falls form 0.15 to 0.10 atm in 25 min . Calculate the average rate of reaction in (a) atm min^(-1) , (b) mol L^(-1) s^(-1)

At 27^@C it was observed during a reaction of hydogenation, that the pressure of hydrogen gas decreasses from 2 atm to 1.1 atm in 75 min. Calculate the rate of reaction (molarity s^(-1) ) (R = 0.0821 litre atm mol^(-1)K^(-1) )

For a 1 molar solution of NaCI in water at 25^(@)C and 1 atm pressure

The pressure of a gas is 2.5 atm . Calculate the value in torr.

If the pressure of H_2 gas is increase from (1) atm to 100 atm. Keeping H^+ concentration constatn at 1 M, the veltage of hydrogen half cell at 25^@ C will be .