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.
All are true except the statement that ions are formed by changing the number protons in an atom’s nucleus.
A neutral atom contains the same number of protons (positive charge) and electrons (negative charge).
If there are <em>more electrons than protons</em>, the atom becomes a <em>negative ion</em>.
If there are <em>fewer electrons than protons</em>, the atom becomes a <em>positive ion</em>.
The protons are in the nucleus, where we can’t easily get at them. The <em>electrons are outside the nucleus</em>, so other chemicals can easily get at them and either remove them or add to their number.
<em>Metals</em> have only a few valence electrons, so it is fairly easy to remove them and <em>form positive ions</em>.
On the left (reactant) side you have 6 N-H bonds and 1 Cl-Cl bond
<span>On the right (product) side you have 4 N-H bonds, 1 N-N bond and 2 H-Cl bonds </span>
<span>Add up the bond energies for each side (multiplying by the number of bonds) and calculate the difference for your ΔHº</span>
Answer:
exothermic entropy is increased
Explanation:
An exothermic process is one whose rate increases when the temperature is decreased. Hence if a decrease in temperature favours the dissolution of more solute at equilibrium, then the process is exothermic.
Similarly, the dissolution of a solute in a solvent increases the disorderliness (entropy) of the system because of the increase in the number of particles present. Hence once a solute in dissolved, the entropy of the system increases.
