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zvonat [6]
3 years ago
8

Consider the reaction. mc032-1.jpg At equilibrium, the concentrations of the different species are as follows. [H2O] = 0.077 M [

Cl2O] = 0.077 M [HClO] = 0.023 M What is the equilibrium constant for the reaction at this temperature?
Chemistry
2 answers:
siniylev [52]3 years ago
8 0
<span>The equilibrium constant is a value which represents the equilibrium of a reaction. It is a reaction quotient when the reaction reached equilibrium. If Keq is greater than 1, the mixture contains mostly the products. On the other hand, if Keq is less than 1, the mixture contains the reactants. It is calculated as follows: 2HClO = H2O + Cl2O

Keq = </span>[H2O] x [Cl2O] / <span>[HClO]^2
Keq = 0.077 x 0.077 / 0.023
Keq = 0.258</span>
SashulF [63]3 years ago
7 0

I just took the test and the answer is 0.089 (A)

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When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant so
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Answer: 54.4 kJ/mol

Explanation:

First we have to calculate the moles of HCl and NaOH.

\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=1.0M\times 0.05=0.05mole

\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=1.0\times 0.05L=0.05mole

The balanced chemical reaction will be,

HCl+NaOH\rightarrow NaCl+H_2O

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.05 mole of HCl neutralizes by 0.05 mole of NaOH

Thus, the number of neutralized moles = 0.05 mole

Now we have to calculate the mass of water:

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = 50ml+50ml=100ml

\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 100ml=100g

Now we have to calculate the heat absorbed during the reaction.

q=m\times c\times (T_{final}-T_{initial})

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q = heat absorbed = ?

c = specific heat of water = 4.18J/g^oC

m = mass of water = 100 g

T_{final} = final temperature of water = 27.5^0C

T_{initial} = initial temperature of metal = 21.0^0C

Now put all the given values in the above formula, we get:

q=100g\times 4.18J/g^oC\times (27.5-21.0)^0C

q=2719.6J=2.72kJ

Thus, the heat released during the neutralization = 2.72 KJ

Now we have to calculate the enthalpy of neutralization per mole of HCl:

0.05 moles of HCl releases heat = 2.72 KJ

1 mole of HCl releases heat =\frac{2.72}{0.05}\times 1=54.4KJ

Thus the enthalpy change for the reaction in kJ per mol of HCl is 54.4 kJ

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