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.
Answer:
No. it,s not possible.
Explanation:
Because you know that ionic compound is having large inter-molecular forces. -77C* is for a compound having v.very small inter-molecular forces.you must know that M.B is directly varing with inter-molecular forces.
Answer:
D. 20.6g
Explanation:
Molarity is defined as ratio between moles of solute (sodium bromide, NaBr) per liter of water.
As density of water is 1g/mL; volume of 400.0g of water is 400.0mL = 0.4000L.
That means 0.400L of 0.500M solution contains:
0.400L × (0.500mol / 1L) = <em>0.200moles of sodium bromide</em>.
In mass (NaBr = 102.9g/mol):
0.200mol NaBr × (102.9g/mol) = 20.6g of NaBr
Right answer is:
<em>D. 20.6g</em>
Unstable nucleus is your answer
Carbon: C
Oxygen: O2
C + O2 —> CO2
Reactants Product
