Answer:
a. Methanol remains the same
b. Methanol decreases
c. Methanol increases
d. Methanol remains the same
e. Methanol increases
Explanation:
Methanol is produced by the reaction of carbon monoxide and hydrogen in the presence of a catalyst as follows; 2H2+CO→CH3OH.
a) The presence or absence of a catalyst makes no difference on the equilibrium position of the system hence the methanol remains constant.
b) The amount of methanol decreases because the equilibrium position shifts towards the left and more reactants are formed since the reaction is exothermic.
c) If the volume is decreased, there will be more methanol in the system because the equilibrium position will shift towards the right hand side.
d) Addition of helium gas has no effect on the equilibrium position since it does not participate in the reaction system.
e) if more CO is added the amount of methanol increases since the equilibrium position will shift towards the right hand side.
Here we have to get the 
of the reaction at 520 K temperature.
The of the reaction is 1.705 atm
We know the relation between and 
is 
, where = The equilibrium constant of the reaction in terms of partial pressure, = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.
Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅
Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole
The given value of is 4.0×10⁻²
The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.
On plugging the values in the equation we get,
Or, = 1.705 atm
Thus, the of the reaction is 1.705 atm
To make a supersaturated solution<span>, make a saturated </span>solution<span> of sugar by adding 360 grams of sugar to 100 mL of water at 80 degrees Celsius. When the water cools back down to 25 degrees, that 360 grams of sugar will still be dissolved even though the water </span>should<span> only dissolve 210 grams of sugar.</span>
