Questions: Question For the following equilibrium: 2 A + B ⇌ C + 2 D If equilibrium concentrations are [A]=1.2 M, [B]=0.75 M, and [C]=1.4 M, and Kc=1.9, what is the equilibrium concentration of D? - Your answer should include two significant figures. Provide your answer below: M

Solution Steps

The equilibrium constant expression for the given reaction is: \[ K_c = \frac{[\mathrm{C}][\mathrm{D}]^2}{[\mathrm{A}]^2[\mathrm{B}]} \]

Given: \[ [\mathrm{A}] = 1.2 \, \text{M}, \quad [\mathrm{B}] = 0.75 \, \text{M}, \quad [\mathrm{C}] = 1.4 \, \text{M}, \quad K_c = 1.9 \]

Substitute these values into the equilibrium constant expression: \[ 1.9 = \frac{(1.4)[\mathrm{D}]^2}{(1.2)^2(0.75)} \]

First, simplify the denominator: \[ (1.2)^2 = 1.44 \] \[ 1.44 \times 0.75 = 1.08 \]

Now, the equation becomes: \[ 1.9 = \frac{1.4[\mathrm{D}]^2}{1.08} \]

Multiply both sides by 1.08 to isolate \([\mathrm{D}]^2\): \[ 1.9 \times 1.08 = 1.4[\mathrm{D}]^2 \] \[ 2.052 = 1.4[\mathrm{D}]^2 \]

Divide both sides by 1.4: \[ [\mathrm{D}]^2 = \frac{2.052}{1.4} \] \[ [\mathrm{D}]^2 = 1.4657 \]

Take the square root of both sides to find \([\mathrm{D}]\): \[ [\mathrm{D}] = \sqrt{1.4657} \approx 1.2107 \]

Round the concentration of D to two significant figures: \[ [\mathrm{D}] \approx 1.2 \, \text{M} \]

Final Answer