In: Chemistry
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
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)