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Question: For the given reversible reaction $2A + B \rightleftharpoons 3C + D$ The expression for the formati...

For the given reversible reaction 2A+B⇌3C+D2A + B \rightleftharpoons 3C + D

The expression for the formation of C is given as: +dCdt=2.4×10−4[A]2[B]−6×10−5[C]3[D]+\frac{dC}{dt} = 2.4 \times 10^{-4} [A]^2[B] - 6 \times 10^{-5}[C]^3[D]

The value of equilibrium constant for the given reaction is (unit of Keq is mol/L for all)

Answer

4

Explanation

Solution

The given reversible reaction is: 2A+B⇌3C+D2A + B \rightleftharpoons 3C + D

The expression for the rate of formation of C is given as: +dCdt=2.4×10−4[A]2[B]−6×10−5[C]3[D]+\frac{dC}{dt} = 2.4 \times 10^{-4} [A]^2[B] - 6 \times 10^{-5}[C]^3[D]

At equilibrium, the net rate of reaction is zero, which means the rate of formation of C is zero: dCdt=0\frac{dC}{dt} = 0

Therefore, at equilibrium: 2.4×10−4[A]2[B]−6×10−5[C]3[D]=02.4 \times 10^{-4} [A]^2[B] - 6 \times 10^{-5}[C]^3[D] = 0

Rearranging the equation, we get: 2.4×10−4[A]2[B]=6×10−5[C]3[D]2.4 \times 10^{-4} [A]^2[B] = 6 \times 10^{-5}[C]^3[D]

The equilibrium constant (KeqK_{eq}) for the given reaction is defined as the ratio of the product concentrations to the reactant concentrations, each raised to the power of their stoichiometric coefficients: Keq=[C]3[D][A]2[B]K_{eq} = \frac{[C]^3[D]}{[A]^2[B]}

From the equilibrium condition derived above, we can express KeqK_{eq}: [C]3[D][A]2[B]=2.4×10−46×10−5\frac{[C]^3[D]}{[A]^2[B]} = \frac{2.4 \times 10^{-4}}{6 \times 10^{-5}}

Now, calculate the value of KeqK_{eq}: Keq=2.4×10−46×10−5K_{eq} = \frac{2.4 \times 10^{-4}}{6 \times 10^{-5}} Keq=2.46×10−410−5K_{eq} = \frac{2.4}{6} \times \frac{10^{-4}}{10^{-5}} Keq=0.4×10(−4−(−5))K_{eq} = 0.4 \times 10^{(-4 - (-5))} Keq=0.4×10(−4+5)K_{eq} = 0.4 \times 10^{(-4 + 5)} Keq=0.4×101K_{eq} = 0.4 \times 10^1 Keq=4K_{eq} = 4

The unit of KeqK_{eq} for this reaction: Δn=(3+1)−(2+1)=4−3=1\Delta n = (3+1) - (2+1) = 4 - 3 = 1. So, the unit is (mol/L)1=mol/L(mol/L)^1 = mol/L, which is consistent with the information given in the question.