Answer
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Hint: According to ideal gas law, partial pressure is inversely proportional to volume. It is directly proportional to the number of moles and temperature. Then the pressure at initial and pressure at equilibrium will be found.
Complete step by step answer:
We have been given the question that a manometer attached to the flask contains ammonia gas.
Then we have been given that ammonia is partially dissociated to nitrogen gas and hydrogen gas as follows
$2N{H_3}(g) \to {N_2}(g) + 3{H_2}(g)$
It is given that ammonia is at $76$ cm Hg pressure. Therefore the initial pressure will be 76 cm Hg
At equilibrium, the pressure will be $76 - 2x,x,3x$ respectively. So the total pressure will be written as $76 - 2x + x + 3x = 76 + 2x$
So the total pressure at equilibrium is $76 + 2x$
We can observe from here that the increase in pressure is $76 - (76 + 2x) = 2x$
It is given that difference in mercury level is six cm
$ \Rightarrow 2x = 6$
$ \Rightarrow x = 3$ cm
So the partial pressure of hydrogen gas $ = 3x$
$ \Rightarrow 3 \times 3 = 9\;cm\;Hg$
So the partial pressure of hydrogen is $9\;cm\;Hg$
So, the correct answer is Option D.
Additional information:
The difference between partial pressure and vapor pressure is that partial pressure is the pressure exerted by the gas individually in the mixture. While vapor pressure is the pressure exerted by the liquid at the thermodynamic equilibrium.
Note: The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.
The partial pressure of gas $ = $ Total pressure $ \times $ mole fraction of gas
$ \Rightarrow P = {P_T}{X_A}$
Also, the total pressure of an ideal gas mixture is the sum of the partial pressure of the gases in the mixture. The pressure exerted by vapors of liquid is called vapor pressure.
Complete step by step answer:
We have been given the question that a manometer attached to the flask contains ammonia gas.
Then we have been given that ammonia is partially dissociated to nitrogen gas and hydrogen gas as follows
$2N{H_3}(g) \to {N_2}(g) + 3{H_2}(g)$
It is given that ammonia is at $76$ cm Hg pressure. Therefore the initial pressure will be 76 cm Hg
At equilibrium, the pressure will be $76 - 2x,x,3x$ respectively. So the total pressure will be written as $76 - 2x + x + 3x = 76 + 2x$
So the total pressure at equilibrium is $76 + 2x$
We can observe from here that the increase in pressure is $76 - (76 + 2x) = 2x$
It is given that difference in mercury level is six cm
$ \Rightarrow 2x = 6$
$ \Rightarrow x = 3$ cm
So the partial pressure of hydrogen gas $ = 3x$
$ \Rightarrow 3 \times 3 = 9\;cm\;Hg$
So the partial pressure of hydrogen is $9\;cm\;Hg$
So, the correct answer is Option D.
Additional information:
The difference between partial pressure and vapor pressure is that partial pressure is the pressure exerted by the gas individually in the mixture. While vapor pressure is the pressure exerted by the liquid at the thermodynamic equilibrium.
Note: The partial pressure of an individual gas is equal to the total pressure multiplied by the mole fraction of that gas.
The partial pressure of gas $ = $ Total pressure $ \times $ mole fraction of gas
$ \Rightarrow P = {P_T}{X_A}$
Also, the total pressure of an ideal gas mixture is the sum of the partial pressure of the gases in the mixture. The pressure exerted by vapors of liquid is called vapor pressure.
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