Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2
when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
= 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value)
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.
c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.
There are several information's already given in the question. Based on the information's the answer can be easily deduced.
We know the formula
<span>P1*V1/T1 = P2*V2/T2
</span>
As the temperature is constant, so T1 and T2 can be negated. The formula changes to
<span>P1*V1 = P2*V2
</span>70 * 1 = 540 * V2
270/540 = V2
<span>V2 = 0.5
</span>
From the above deduction, we can conclude that the new volume is 0.5 liters. I hope that the procedure is clear enough for you to understand.
The first ionisation energy of silicon is greater than that of phosphorus.
Answer:
Explanation:
The SI unit for mass is the KG
