A 1.00 mol sample of A was placed in a 2.00 L reactor and heated to 342 ${ } ^ { \circ } \mathrm { C }$ until the system reached equilibrium. The contents of the reactor then contained 0.0488 mol of C. What is K for the reaction 2A(g) $f$ 2B(g) +C(g) at the same temperature?

Question

Quizplus · Accepted Answer

First, we need to write out the balanced chemical equation and the expression for K:
$$2A(g) ightleftharpoons 2B(g) + C(g)$$
$$K = \frac{\left[ B ight] ^ 2 \cdot \left[ C ight]}{\left[ A ight] ^ 2}$$

At equilibrium, let's assume that the amount of A that reacts is "x" mol. Therefore, the equilibrium concentrations of A, B, and C are:

$$[\mathrm{A}] = (1.00 - x)\,\mathrm{M}$$
$$[\mathrm{B}] = x\,\mathrm{M}$$
$$[\mathrm{C}] = 0.0488\,\mathrm{M}$$

From the equation, we see that 2 moles of B are formed for every 2 moles of A that react. Therefore, the equilibrium concentration of B is also "x" mol/L.

Substituting these values into the expression for K, we get:

$$K = \frac{\left( x\,\mathrm{M} ight) ^ 2 \cdot \left( 0.0488\,\mathrm{M} ight)}{\left( 1.00 - x)\,\mathrm{M} ight) ^ 2}$$

We can solve for "x" by substituting the known values into the expression and solving the resulting quadratic equation. Using the quadratic formula, we get:

$$x = 0.226\,\mathrm{M}$$

Finally, we substitute this value back into the expression for K to get:

$$K = \frac{\left( 0.226\,\mathrm{M} ight) ^ 2 \cdot \left( 0.0488\,\mathrm{M} ight)}{\left( 1.00 - 0.226\,\mathrm{M} ight) ^ 2} = 2.85 	imes 10 ^ { - 4 }$$

Therefore, the answer is D.

A 100 Mol Sample of a Was Placed in a 2