The value of ka for nitrous acid (hno2) at 25 ∘c is 4.5×10−4 .a. write the chemical equation for the equilibrium that corresponds to ka1. h+(aq)+no−2(aq)⇌hno2(aq)2. hno2(aq)⇌h+(aq)+no−2(aq)3. hno2(aq)⇌h−(aq)+no+2(aq)4. hno2(aq)+h+(aq)⇌h2no+2(aq)5. hno2(aq)+h−(aq)⇌h2no+2(aq)b. by using the value of ka , calculate δg∘ for the dissociation of nitrous acid in aqueous solution. c. what is the value of δg at equilibrium? d. what is the value of δg when [h+] = 5.9×10−2 m , [no−2] = 6.7×10−4 m , and [hno2] = 0.21 m ?

Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

   ΔGº= - RT ln Ka

   ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)

  ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

Q= 1.88 10⁻⁴

We know that   ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

ΔG= -2162.4 J/mol

Notice that ΔG<0, so the process is spontaneous in that direction.

these nuts