Answer:
The more spread out their energy becomes
Answer:
So, 10 mole of water will weigh (18x10) = 180g.
It is called phosphorescent substances.
3 L will be the final volume for the gas as per Charle's law.
Answer:
Explanation:
The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.
V∝ T
Since, here two volumes are given and at two different temperatures with constant pressure. Then as per Charle's law, the relation between the volumes of air at different temperature will be
So in this case, V1 = 6 L and T1 = 80° C. Similarly, T2 = 40° C. So we have to determine the V2.
So, 3 L will be the final volume for the gas as per Charle's law.
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.
