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AlekseyPX
3 years ago
5

Consider the following system at equilibrium:

Chemistry
1 answer:
alexgriva [62]3 years ago
6 0

Answer:

A - Increase (R), Decrease (P), Decrease(q), Triple both (Q) and (R)

B - Increase(P), Increase(q), Decrease (R)

C - Triple (P) and reduce (q) to one third

Explanation:

<em>According to Le Chatelier principle, when a system is in equilibrium and one of the constraints that affect the rate of reaction is applied, the equilibrium will shift so as to annul the effects of the constraint.</em>

P and Q are reactants, an increase in either or both without an equally measurable increase in R (a product) will shift the equilibrium to the right. Also, any decrease in R without a corresponding decrease in either or both of P and Q will shift the equilibrium to the right. Hence, Increase(P), Increase(q), and Decrease (R) will shift the equilibrium to the right.

In the same vein, any increase in R without a corresponding increase in P and Q will shift the equilibrium to the left. The same goes for any decrease in either or both of P and Q without a counter-decrease in R will shift the equilibrium to the left. Hence, Increase (R), Decrease (P), Decrease(q), and Triple both (Q) and (R) will shift the equilibrium to the left.

Any increase or decrease in P with a commensurable decrease or increase in Q (or vice versa) with R remaining constant will create no shift in the equilibrium. Hence, Triple (P) and reduce (q) to one third will create no shift in the equilibrium.

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What is the concentration (in M) of a 225ml potassium sulfate solution that contains 4.15g of potassium?
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The concentration of solution in M or mol/L can be calculated using the following formula:

C=\frac{n}{V} .... (1)

Here, n is number of moles and V is volume of solution in L.

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1 mol of potassium will be there in 0.5 mol of potassium sulfate.

Mass of potassium is 4.15 g, molar mass is 39.1 g/mol.

Number of moles can be calculated as follows:

n=\frac{m}{M}

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Putting the values,

n=\frac{(4.15 g}{(39.1 g/mol}=0.1061 mol

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C=\frac{(0.053 mol}{0.225 L}=0.236 M

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