Question

A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration of NO in the vessel is 0.161M. At equilibrium the vessel also contains N2, H2O, and H2. What is the value of the equilibrium constant for Kc for the following reaction?
2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)

Answer 1

Answer:

The equilibrium constant Kc for this reaction is 19.4760

Explanation:

The volume of vessel used= $100$ ml

Initial moles of NO= $\frac{2.60}{10^2}$ moles

Initial moles of H2= $\frac{1.30}{10^2}$ moles

Concentration of NO at equilibrium= $0.161$M

$Concentration(in M)=\frac{moles}{volume(in litre)}$

Moles of NO at equilibrium= $0.161(\frac{100}{1000})$

                                            =$\frac{1.61}{10^2}$ moles

               

                    2H2 (g)        +    2NO(g) <—>    2H2O (g) +    N2 (g)

Initial          :1.3*10^-2          2.6*10^-2                0                   0        moles

Equilibrium:1.3*10^-2 - x     2.6*10^-2-x              x                   x/2     moles

∴$\frac{2.60}{10^2}-x=\frac{1.61}{10^2}$

⇒$x=\frac{0.99}{10^2}$

$Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)$

Equilibrium:0.31*10^-2      1.61*10^-2          0.99*10^-2        0.495*10^-2  moles

⇒$Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2} (0.1)$

⇒$Kc=19.4760$