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trasher [3.6K]
3 years ago
12

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

Chemistry
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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Kemmi pipets 25.00 ml of pure 1-propanol (c3h7oh, a liquid organic alcohol) into a 100.0 ml volumetric flask. she dilutes it wit
algol13
Q1)
As Kemmi pipettes a volume of 25.00 ml of the solution
density of pure propanol is 0.803 g/ml
This means that in 1000 ml of solution - 0.803 g of pure propanol
Therefore in 25.00 ml of solution - 0.803 g x 25.00 ml / 1000 ml
                                                     = 0.0201 g
Using molar mass, number of moles can be calculated= 0.0201 g / 60.09 g/mol
                                                                                       = 3.35 x 10⁻⁴ mol
therefore the number of pure propanol moles in exactly 25.00 ml is
3.35 x 10⁻⁴ mol

Q2)
molarity is the concentration of the solution. It can be defined as the number of moles of solute per liter of solution
we know the number of moles in 25.00 ml of solution. When its diluted in a 100.00 ml volumetric flask, number of moles remain constant but now the volume over which the moles of solute are dissolved is increased.
therefore number of moles = 3.35 x 10^(-4) mol
volume over which its dissolved - 100.00 / 10³ dm³
                                                    = 1.0000 x10⁻¹ dm³
the molarity = 3.35 x 10⁻⁴ mol / 1.0000 x10⁻¹ dm³
                    = 3.35 x 10⁻³ mol/dm³
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