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Chemistry Question on Equilibrium

In a closed cylinder of capacity 24.6 L24.6\,L, the following reaction occurs at 27∘C27^{\circ} C. A2(s)⇌B2(s)+2C(g)A_{2}(s) \rightleftharpoons B_{2}(s)+2 C(g) At equilibrium, 1 g1 \,g of B2(s)B_{2}(s) (molar mass =50 g mol−1=50\, g\, mol ^{-1} ) is present. The equilibrium constant KpK_{p} for the equilibrium in atm 2^{2} unit is (R=0.082 L atm K−1 mol−1)\left(R=0.082\, L\, atm \,K ^{-1} \,mol ^{-1}\right)

A

1.6×10−21.6 \times 10^{-2}

B

1.6×10−51.6 \times 10^{-5}

C

1.6×10−31.6 \times 10^{- 3}

D

1.6×10−41.6 \times 10^{-4}

Answer

1.6×10−31.6 \times 10^{- 3}

Explanation

Solution

Equilibrium concentration of B2B_{2},

[B2]equilibrium = mass  molar mass × volume \left[B_{2}\right]_{\text {equilibrium }} =\frac{\text { mass }}{\text { molar mass } \times \text { volume }}
=150×24.0=8.13×10−4=\frac{1}{50 \times 24.0}=8.13 \times 10^{-4}

For the reaction,


∴[C]=2×8.13×10−4=1.63×10−3\therefore[C] =2 \times 8.13 \times 10^{-4}=1.63 \times 10^{-3}
Kc=[C]2=[1.63×10−3]2=2.66×10−6K_{c} =[C]^{2}=\left[1.63 \times 10^{-3}\right]^{2}=2.66 \times 10^{-6}

Again from, Kp=KC(KT)ΔnyK_{p}=K_{C}(K T)^{\Delta n_{y}}

Here, Δng=2−0=2\Delta n_{g}=2-0=2
Kp=2.66×10−6×(0.082×300)2K_{p} =2.66 \times 10^{-6} \times(0.082 \times 300)^{2}
≈1.6×10−3\approx 1.6 \times 10^{-3}