A. The reaction would remain in equilibrium
<h3>Further explanation</h3>
Given
Reaction
H₂ + I₂ ⇔ 2HI
Required
the effect of pressure changes
Solution
In the equilibrium system :
<em>Reaction = - action
</em>
⇒shift the reaction to the right or left.
The pressure usually affects the gas equilibrium system(only count the number of moles of gases)
The addition of pressure, the reaction will shift towards a smaller reaction coefficient ((the fewest moles of gas )
Reaction
H₂ + I₂ ⇔ 2HI
The reactant side of the equation has 2 moles of a gas(1 mole H₂ and 1 mole I₂) ; the product side has 2 moles HI
So the total number of moles from both sides is the same(2 moles) so that the change in volume (pressure) <em>does not change the direction of equilibrium⇒No shift will occur
</em>
Because area of the container has increased , there will be fewer of collisions per unit area and the pressure will decrease . Volume is inversely proportional to pressure , if the number of particles and temperature is constant
( V = 1/P) and number of particle is proportional to pressure if average Kinetic energy of the particle remain same , the average force particle will remain same too so at some places and there will be more collision and there is greater pressure
2Ca + O2 = 2CaO
First, determine which is the excess reactant
72.5 g Ca (1 mol) =1.8089725036
(40.078 g)
65 g O2 (1 mol) =2.0313769611
(15.999g × 2)
Since the ratio of to O2 is 2:1 in the balanced reaction, divide Ca's molar mass by 2 to get 0.9044862518. this isn't necessary because Ca is already obviously the limiting reactant. therefore, O2 is the excess reactant.
Now do the stoichiometry
72.5 g Ca (1 mol Ca) (1 mol O2)
(40.078 g Ca)(2 mol Ca)(31.998g O2)
=0.0282669621 g of O2 left over
