Question

The equilibrium constant K\(_c\) at 298 K for the reaction A + B \(\rightleftharpoons\) C + D is 100. Starting with an equimolar solution with concentrations of A, B, C and D all equal to 1 M, the equilibrium concentration of D is ___________ \(\times\) 10\(^{-2}\) M. (Nearest integer)

Hint:

Always calculate the reaction quotient \(Q_c\) first when initial concentrations of all species are given. This tells you whether the reaction will shift forwards (\(Q<K\)) or backwards (\(Q>K\)), which is crucial for correctly setting up the 'Change' row in the ICE table (i.e., whether 'x' is added or subtracted).

Answer 1

Correct Answer: 182

Step 1: Understanding the Question
We are given a reversible reaction with its equilibrium constant \(K_c\). We start with a non-equilibrium mixture and need to find the concentration of one of the products, D, once equilibrium is reached.
Step 2: Key Formula or Approach
First, we calculate the reaction quotient \(Q_c\) to determine the direction the reaction will shift to reach equilibrium. Then, we use an ICE (Initial, Change, Equilibrium) table to set up an expression for \(K_c\) in terms of the change in concentration, and solve for it.
Step 3: Detailed Calculation
Calculate the initial Reaction Quotient (\(Q_c\)):
The reaction is A + B \(\rightleftharpoons\) C + D.
Initial concentrations: [A]\(_i\) = [B]\(_i\) = [C]\(_i\) = [D]\(_i\) = 1 M.
\[ Q_c = \frac{[\text{C}]_i[\text{D}]_i}{[\text{A}]_i[\text{B}]_i} = \frac{(1)(1)}{(1)(1)} = 1 \] Determine the direction of shift:
We are given \(K_c = 100\). Since \(Q_c (1)<K_c (100)\), the reaction will proceed in the forward direction to reach equilibrium. This means the concentrations of C and D will increase, and A and B will decrease.
Set up an ICE table:
Let 'x' be the change in concentration in M.
\begin{tabular}{l c c c c c c c} & A & + & B & \(\rightleftharpoons\) & C & + & D
Initial (I) & 1 & & 1 & & 1 & & 1
Change (C) & -x & & -x & & +x & & +x
Equilibrium (E) & 1-x & & 1-x & & 1+x & & 1+x
\end{tabular}
Solve for x using the K\(_c\) expression:
\[ K_c = \frac{[\text{C}]_{eq}[\text{D}]_{eq}}{[\text{A}]_{eq}[\text{B}]_{eq}} = \frac{(1+x)(1+x)}{(1-x)(1-x)} = \left(\frac{1+x}{1-x}\right)^2 \] \[ 100 = \left(\frac{1+x}{1-x}\right)^2 \] Taking the square root of both sides:
\[ 10 = \frac{1+x}{1-x} \] \[ 10(1-x) = 1+x \] \[ 10 - 10x = 1 + x \] \[ 9 = 11x \] \[ x = \frac{9}{11} \approx 0.8181 \text{ M} \] Calculate the equilibrium concentration of D:
[D]\(_{eq}\) = 1 + x = 1 + \(\frac{9}{11}\) = \(\frac{20}{11} \approx 1.8181 \text{ M}\).
Express the answer in the required format:
[D]\(_{eq}\) = 1.8181 M = 181.81 \(\times\) 10\(^{-2}\) M. Step 4: Final Answer
Rounding to the nearest integer, the value is 182.