Question

A 100 mL aqueous soution containing 0.1 moles of HCN and 0.1 moles of NH₃ is allowed to react in a specialized calorimeter. The reaction proceeds until it reaches equilibrium. Over the course of this reaction, enough heat is released to warm 25g of pure water by 2.64 °C. Given the specific heat of water (4.184 J/(g*K)) and the reaction below, calculate the equilibrium constant for this reaction.

HCN(aq) + NH₃(aq) → NH₄+(aq)+ CN-(aq)        ΔH = -5.4 kJ/mol

Answer 1

HCN(aq) + NH3(aq) → NH4+(aq) + CN-(aq)

initial

No of mole of HCN = 0.1/0.1 = 1 M

No of mole of NH3 = 0.1/0.1 = 1 M

at equilibrium

q = m*s*DT

   = 25*4.18*2.64

   = 275.88 joule.

   = 0.275 kj

1 mol = 5.4 kj/mol

x =   0.275 kj

x = 0.275/ 5.4   = 0.06 mol

so that, No of mole of NH4+ produced = 0.06 mol

concentration of NH4+ = 0.06 / 0.1 = 0.006 M

concentration of CN- = 0.006 M

concentration of HCN = 1-0.006 = 0.994 M

concentration of NH3 = 0.994 M

kc = [nh4+][cn-]/[HCN][NH3]

   = 0.006^2 / 0.994^2

   = 3.64*10^(-5)