How many moles of ammonia gas occupy 50 mL at at 700 kPa and 30.0 0C?
1 answer:
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Answer:
The new equilibrium total pressure will be increased to one-half to initial total pressure.
Explanation:
From the information given :
The equation of the reaction can be represented as;
From above equation:
2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide
So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.
So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.
Let the total pressure at the initial equilibrium be
and the total pressure at the final equilibrium be
According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Thus;
P ∝ 1/V
P = K/V
PV = K
where K = constant
So;
PV = constant
Hence;
From the foregoing; since the volume is decreased to one- half to initial Volume; then ,
also;
Thus ;
Dividing both sides by
From ;
Thus; The new equilibrium total pressure will be increased to one-half to initial total pressure.
Answer:
Kc = 1.54e - 31 / 2.61e - 24
Explanation:
1 ) ; Kc = 1.54e - 31
2) ; Kc = 2.16e - 24
upon reversing ( 2 ) equation
Kc = 1/2.16e - 24
now adding 1 and reversed equation (2)
we get ,
Kc = 1.54e-31 × 1/2.61e - 24
equilibrium constant of equation (3) is -
Kc = 1.54e - 31 / 2.61e - 24