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3 years ago

12

For an endothermic reaction, how will the value for Keq change when the temperature is increased?

2 answers:

Nataly [62]3 years ago

4 0

I put the answer <em>C: Keq will increase</em>, on PLATO. Hope this works for you!

Vlada [557]3 years ago

3 0

Answer:

C. Keq will increase

Explanation:

Endothermic means the reaction will <em>absorb heat</em> to occur, hence an increase in temperature will favor the reaction, increasing the formation of products.

Keq is the equilibrium constant of the reaction, and it can be defined as:

$Keq = \frac{concentration of products}{concentration of reactants}$

In an endothermic reaction the equilibrium will be shift towards the formation of products when the temperature is increased, the concentration of products will increase (in relation to the concentration of reactants) ; so as a result the equilibrium constant Keq will also increase.

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1 year ago

The solubility of oxygen gas in water at 40 ∘c is 1.0 mmol/l of solution. What is this concentration in units of mole fraction?

juin [17]

The formula for mole fraction is:

$mole fraction of solute = \frac{number of moles of solute}{total number of moles of solution}$ -(1)

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1.0 mmol/L means 1.0 mmol are present in 1 L.

Converting mmol to mol:

$1.00 mmol\times \frac{1 mol}{1000 mmol} = 0.001 mol$

So, moles of oxygen = 0.001 mol

For moles of water:

1 L of water = 1000 mL of water

Since, the density of water is 1.0 g/mL.

$Density = \frac{mass}{volume}$

$Mass = 1.0 g/ml\times 1000 mL = 1000 g$

So, the mass of water is 1000 g.

Molar mass of water = 18 g/mol.

Number of moles of water = $\frac{1000 g}{18 g/mol} = 55.55 mol$

Substituting the values in formula (1):

$mole fraction = \frac{0.001}{55.55+0.001}$

$mole fraction = 1.8\times 10^{-5}$

Hence, the mole fraction is $1.8\times 10^{-5}$ .

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3 years ago