Consider 2 nocl(g) \longleftrightarrow⟺ 2 no(g) + cl2 (g) at 25oc under conditions other than equilibrium, there are 1.20 moles of nocl , 0.450 moles of no, and 0.220 moles of cl2 in a 5.00 l flask. kc = 1.86 x 10-1. show work for credit. a. calculate q. b. predict the direction of the reaction. c. calculate equilibrium concentrations of all species present. the equilibrium concentration of cl2 is 0.0490 m.

a) Q = 6.1875x10⁻³

b) The direction of the reaction is to form the products.

c) [Cl₂]e = 0.094 M

[NO]e =  0.190 M

[NOCl]e = 0.140 M

Explanation:

a) Q is the reaction quotient, and for a generic reaction aA + bB ⇄ cC + dD it is

Which is the same equation for Kc, but in Kc expressions, the concentrations are in the equilibrium. Q is calculated at any time. So, for the reaction given

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

[Cl₂] = 0.220/5.00 = 0.044 M

[NO] = 0.450/5.00 = 0.090 M

[NOCl] = 1.20/5.00 = 0.240 M

Q = (0.044)x(0.090)²/(0.240)²

Q = 6.1875x10⁻³

b) Q < Kc, which means that there are fewer products to what are needed to the equilibrium. So the direction of the reaction is to form the products.

c)

2NOCl(g) ⇄ 2NO(g) + Cl2(g)

0.240           0.090      0.044         Initial

-2x                 +2x          +x             Reacts (stoichiometry is 2:2:1)

0.240-2x   0.090+2x   0.044+x     Equilibrium

3.564x10⁻⁴+0.01584x+0.176x²+8.1x10⁻³x+0.36x²+4x³ = 0.010714-0.17856x+0.744x²

4x³ - 0.208x² + 0.2025x - 0.01036 = 0

Solving this third grade equation in a computer program:

x = 0.05 M

So:

[Cl₂]e = 0.044 + 0.05 = 0.094 M

[NO]e = 0.090 + 2x0.05 = 0.190 M

[NOCl]e = 0.240 - 2x0.05 = 0.140 M

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g