Math  /  Algebra

QuestionAt 700 K , the reaction 2SO2( g)+02 \mathrm{SO}_{2}(\mathrm{~g})+0 2( g)2SO3( g)2(\mathrm{~g}) \leftrightharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}) has the equilibrium constant Kc=4.3×106K_{c}=4.3 \times 10^{6}, and the following concentrations are present: [SO2]=0.10M;[SO3]=10.M;[O2]\left[\mathrm{SO}_{2}\right]=0.10 \mathrm{M} ;\left[\mathrm{SO}_{3}\right]=10 . \mathrm{M} ;\left[\mathrm{O}_{2}\right] =0.10M=0.10 \mathrm{M}. Which of the following is true based on the above? Qc>Kc\mathrm{Q}_{c}>\mathrm{K}_{c}, the reaction proceeds from left to right to reach equilibrium Qc<Kc\mathrm{Q}_{c}<K_{c}, the reaction proceeds from left to right to reach equilibrium Qc<KcQ_{c}<K_{c}, the reaction proceeds from right to left to reach equilibrium Qc=KcQ_{c}=K_{c}, the reaction is currently at equilibrium Qc>Kc\mathrm{Q}_{c}>\mathrm{K}_{c}, the reaction proceeds from right to left to reach equilibrium

Studdy Solution

STEP 1

What is this asking? We need to figure out if this reaction is at equilibrium, and if not, which way it will shift to get there! Watch out! Don't mix up KcK_c and QcQ_c! KcK_c is the equilibrium constant, while QcQ_c is the reaction quotient – it tells us where we are *right now*.

STEP 2

1. Calculate QcQ_c
2. Compare QcQ_c and KcK_c

STEP 3

Remember, QcQ_c is just like KcK_c, but with current concentrations instead of equilibrium concentrations.
For the reaction 2SO2( g)+O2( g)2SO3( g)2 \mathrm{SO}_{2}(\mathrm{~g})+ \mathrm{O}_{2}(\mathrm{~g}) \leftrightharpoons 2 \mathrm{SO}_{3}(\mathrm{~g}), the expression is: Qc=[SO3]2[SO2]2[O2]Q_c = \frac{[\mathrm{SO}_3]^2}{[\mathrm{SO}_2]^2 \cdot [\mathrm{O}_2]}

STEP 4

We're given [SO2]=0.10[\mathrm{SO}_2] = 0.10 M, [SO3]=10[\mathrm{SO}_3] = 10 M, and [O2]=0.10[\mathrm{O}_2] = 0.10 M.
Let's plug these values into our QcQ_c expression: Qc=(10)2(0.10)2(0.10)Q_c = \frac{(10)^2}{(0.10)^2 \cdot (0.10)}

STEP 5

Time to crunch the numbers! Qc=1000.010.10=1000.001=100,000=1.0×105Q_c = \frac{100}{0.01 \cdot 0.10} = \frac{100}{0.001} = 100,000 = 1.0 \times 10^5 So, our **reaction quotient**, QcQ_c, is 1.0×1051.0 \times 10^5.

STEP 6

We're told that Kc=4.3×106K_c = 4.3 \times 10^6.

STEP 7

Now, let's see how QcQ_c and KcK_c stack up.
We have Qc=1.0×105Q_c = 1.0 \times 10^5 and Kc=4.3×106K_c = 4.3 \times 10^6.
Clearly, QcQ_c is smaller than KcK_c (Qc<KcQ_c < K_c).

STEP 8

Since Qc<KcQ_c < K_c, the reaction needs to proceed from **left to right** to reach equilibrium.
This means more products will be formed to increase the numerator of QcQ_c and decrease the denominator until QcQ_c equals KcK_c.

STEP 9

Qc<KcQ_c < K_c, and the reaction proceeds from left to right to reach equilibrium.

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