Equilibrium

A reversible reaction where rate of the forward reaction equals to the rate of the backward reaction is called the equilibrium. It results in constant reactants and products concentrations over time.

On increasing pressure, equilibrium moves in that direction where number of gaseous moles decreases.

Kindly go through the solution

K_h = Kw/Ka = 2.22 x 10^-5

pH =   $\frac{1}{2}\left[p{K}_w+p{K}_a-p{K}_b\right]$

$=\frac{1}{2}\left[14+4.75-5.23\right]$

$=6.76\cong 7$          

It happens in reversible reactions when the rate of the forward reaction becomes equal to the rate of the backward reaction. Result in the same concentration of reactants and product.

If the conditions of equilibrium are changed, it shifts to oppose the change. For example, in Haber's process, high pressure favors NH? formation.

The equilibrium constant is the ratio of the concentrations of products to reactants, each raised to the power of their stoichiometric coefficients. For reversible reactions.

Here, total meq of acetic acid = 50 × 0.1 = 5

And total meq of NaOH = 25 × 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH₃COONa = 2.5

$pH=pKa+lo{g}_{10}\frac{\left[S\right]}{\left[A\right]}$

$pH=4.76+lo{g}_{10}\frac{2.5}{2.5}=4.76=476\times 10^{-2}$  

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

  $P_{Total}=\frac{nRT}{V}=\frac{9\times 0.0821\times 610}{100}=4.5atm$

  $P_{PC{l}_5}=X_{PC{l}_5}\times P_T=\frac{5}{9}\times 4.5=2.5atm$

  $P_{Ar}=X_{Ar}\times P_T=\frac{4}{9}\times 4.5=2atm$

  $PC{l}_5\stackrel{}{?}PC{l}_3+C{l}_2$

2.5                       0            0&

Moles of PCl₅ = 5 mol

Moles of Ar = 4 mol

Total no of moles = 9 moles

$P_{Total}=\frac{nRT}{V}=\frac{9\times 0.0821\times 610}{100}=4.5atm$  

$P_{PC{l}_5}=X_{PC{l}_5}\times P_T=\frac{5}{9}\times 4.5=2.5atm$  

$P_{Ar}=X_{Ar}\times P_T=\frac{4}{9}\times 4.5=2atm$  

$PC{l}_5\stackrel{}{?}PC{l}_3+C{l}_2$  

2.5                       0            0&