Each of the following reactions were in equilibrium when the pressure of their containers was doubled. Chose which way the reaction shifted after the pressure change: 2NH3 (g)㈠No(g) +3H2(g) [Select ]
2Na3 PO4 (aq) + 3CaCl2 (aq) ㈠ Ca3 (PO4 )2 (s) + 6NaCl(aq) [Select ]
2C0(g)+O2 (g)->2CO2 (g) [Select ]
2H1(g) ㈠ H2(g) + 12(g) Neither

Each of the following reactions was in equilibrium when the pressure of their containers was doubled. Chose which way the reaction shifted after the pressure change:

2NH3 (g)->N2(g) +3H2(g)

2Na3 PO4 (aq) + 3CaCl2 (aq) -> Ca3 (PO4 )2 (s) + 6NaCl(aq)  

2CO(g)+O2 (g)->2CO2 (g)  

2HI(g) -> H2(g) + I2(g)

Explanation:

Effect of pressure on equilibrium:

When pressure is increased on an equilibrium system,then equilibrium will shift in such a direction towards less number of moles of substrates.

For the first system,

2NH3 (g)->N2(g) +3H2(g)

increase in pressure,shifts the equilibrium towards the left side that is the formation of ammonia is favored.

For the second reaction:

2Na3 PO4 (aq) + 3CaCl2 (aq) -> Ca3 (PO4 )2 (s) + 6NaCl(aq)

The quilibrium will shift towards right.

Becuase right side less number of moles of substrates are there.

2CO(g)+O2 (g)->2CO2 (g)

For this system,the equilibrium will shift towards right side that is formation of CO2 gas is favored.

For the last system,

2HI(g) -> H2(g) + I2(g)

there is no effect of pressure.

Becuase the number of moles of substrates are same on both sides.

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8