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.
The equilibrium expression shows the ratio between
products and reactants. This expression is equal to the concentration of the
products raised to its coefficient divided by the concentration of the
reactants raised to its coefficient. The correct equilibrium expression for the
given reaction is:<span>
<span>H2CO3(aq) + H2O(l) = H3O+(aq) + HCO3-1(aq)
Kc = [HCO3-1] [H3O+] / [H2O] [H2CO3]</span></span>
Answer: The volume of 0.640 grams of
gas at Standard Temperature and Pressure (STP) is 0.449 L.
Explanation:
Given: Mass of
gas = 0.640 g
Pressure = 1.0 atm
Temperature = 273 K
As number of moles is the mass of substance divided by its molar mass.
So, moles of
(molar mass = 32.0 g/mol) is as follows.

Now, ideal gas equation is used to calculate the volume as follows.
PV = nRT
where,
P = pressure
V = volume
n = no. of moles
R = gas constant = 0.0821 L atm/mol K
T = temperature
Substitute the values into above formula as follows.

Thus, we can conclude that the volume of 0.640 grams of
gas at Standard Temperature and Pressure (STP) is 0.449 L.