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Tom [10]
3 years ago
5

Pre-laboratory Questions 1. Using LeChâtelier’s principle, determine whether the reactants or products are favored and the direc

tion of the shift in equilibrium (left or right) in the following examples. 2CO(g) + O2(g)  2CO2(g) + heat a. Remove O2 b. Lower the temperature c. Add CO d. Remove CO2 e. Decrease pressure
Chemistry
1 answer:
3241004551 [841]3 years ago
8 0
I think the answer is C but don’t quote me on that
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Read 2 more answers
At a certain temperature, 0.760 mol SO3 is placed in a 1.50 L container. 2SO3(g) = 2SO2(g) + O2(g)At equilibrium, 0.130 mol O2 i
olganol [36]

Answer:

K_c=0.0867

Explanation:

Moles of SO₃ = 0.760 mol

Volume = 1.50 L

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.760}{1.50\ L}

[SO₃] = 0.5067 M

Considering the ICE table for the equilibrium as:

\begin{matrix} & 2SO_3_{(g)} & \rightleftharpoons & 2SO_2_{(g)} & + & O_2_{(g)}\\At\ time, t = 0 & 0.5067 &&0&&0 \\ At\ time, t=t_{eq} & -2x &&2x&&x \\----------------&-----&-&-----&-&-----\\Concentration\ at\ equilibrium:- &0.5067-2x&&2x&&x\end{matrix}

Given:    

Equilibrium concentration of  O₂ = 0.130 mol

Volume = 1.50 L

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.130}{1.50\ L}

[O₂] = x = 0.0867 M

[SO₂] = 2x = 0.1733 M

[SO₃] = 0.5067-2x = 0.3334 M

The expression for the equilibrium constant is:

K_c=\frac {[SO_2]^2[O_2]}{[SO_3]^2}  

K_c=\frac{(0.3334)^2\times 0.0867}{(0.3334)^2}

K_c=0.0867

5 0
3 years ago
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