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3 years ago

Consider the reaction: H2(g) + Cl2(g) --> 2HCl(g)

Chemistry

1 answer:

suter [353]3 years ago

6 0

Answer : The entropy change of reaction for 2.28 moles of $H_2$ reacts at standard condition is 45.8 J/K

Explanation :

The given balanced reaction is,

$H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$

The expression used for entropy change of reaction $(\Delta S^o)$ is:

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{HCl}\times \Delta S_f^0_{(HCl)}]-[n_{H_2}\times \Delta S_f^0_{(H_2)}+n_{Cl_2}\times \Delta S_f^0_{(Cl_2)}]$

where,

$\Delta S^o$ = entropy change of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

$\Delta S_f^0_{(H_2)}$ = 130.684 J/mol.K

$\Delta S_f^0_{(Cl_2)}$ = 223.066 J/mol.K

$\Delta S_f^0_{(HCl)}$ = 186.908 J/mol.K

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (186.908J/K.mole)]-[1mole\times (130.684J/K.mole)+1mole\times (223.066J/K.mole)}]$

$\Delta S^o=20.066J/K$

Now we have to calculate the entropy change of reaction for 2.28 moles of $H_2$ reacts at standard condition.

From the reaction we conclude that,

As, 1 moles of $H_2$ has entropy change = 20.066 J/K

So, 2.28 moles of $H_2$ has entropy change = $\frac{2.28}{1}\times 20.066=45.8J/K$

Therefore, the entropy change of reaction for 2.28 moles of $H_2$ reacts at standard condition is 45.8 J/K

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You place 36.5 ml of 0.266 M Ba(OH)2 in a coffee-cup calorimeter at 25.00°C and add 56.6 ml of 0.648 M HCl, also at 25.00°C. Aft

olya-2409 [2.1K]

Answer : The enthalpy of reaction $(\Delta H_{rxn})$ is, -96.9 kJ/mole

Explanation :

First we have to calculate the mass of solution.

$Mass=Density\times Volume$

Volume of solution = Volume of HCl + Volume of $Ba(OH)_2$

Volume of solution = 56.6 mL + 36.5 mL

Volume of solution = 93.1 mL

Density of solution = 1 g/mL

$Mass=1g/mL\times 93.1mL=93.1g$

The mass of solution is, 93.1 grams.

Now we have to calculate the heat released in the system.

Formula used :

$Q=m\times c\times \Delta T$

or,

$Q=m\times c\times (T_2-T_1)$

where,

Q = heat released = ?

m = mass = 93.1 g

$C_p$ = specific heat capacity of water = $4.184J/g^oC$

$T_1$ = initial temperature = $25.0^oC$

$T_2$ = final temperature = $29.83^oC$

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

$Q=93.1g\times 4.184J/g^oC\times (29.83-25.00)^ioC$

$Q=1881.43J=1.88kJ$ (1 kJ = 1000 J)

Now we have to calculate the moles of $Ba(OH)_2$ and HCl.

$\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}$

$\text{Moles of }Ba(OH)_2=0.266M\times 0.0365L=9.71\times 10^{-3}mol$

and,

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}$

$\text{Moles of }HCl=0.648M\times 0.0566L=3.66\times 10^{-2}mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O$

From the balanced reaction we conclude that

As, 1 mole of $Ba(OH)_2$ react with 2 mole of $HCl$

So, $9.71\times 10^{-3}$ moles of $Ba(OH)_2$ react with $9.71\times 10^{-3}\times 2=0.0194$ moles of $HCl$

From this we conclude that, $HCl$ is an excess reagent because the given moles are greater than the required moles and $Ba(OH)_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $H_2O$

From the reaction, we conclude that

As, 1 mole of $Ba(OH)_2$ react to give 2 mole of $H_2O$

So, $9.71\times 10^{-3}$ moles of $Ba(OH)_2$ react with $9.71\times 10^{-3}\times 2=0.0194$ moles of $H_2O$

Now we have to calculate the change in enthalpy of the reaction.

$\Delta H_{rxn}=-\frac{q}{n}$

where,

$\Delta H_{rxn}$ = enthalpy of reaction = ?

q = heat of reaction = 1.88 kJ

n = moles of reaction = 0.0194 mole

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

$\Delta H_{rxn}=-\frac{1.88kJ}{0.0194mole}=-96.9kJ/mole$

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction $(\Delta H_{rxn})$ is, -96.9 kJ/mole

3 0

3 years ago

When OH- and H+ ions are brought together, they quickly combine to form water molecules.

Alik [6]

True because positive and negative ions makes neutral hence the water is neutral

5 0

3 years ago

What is isotope <br>the percentage of the solid components of soil

zubka84 [21]

Answer:

an isotope consists of two or more forms of the same elements that contains equal number of protons but different number of neutrons in their nuclei but differ in relative atomic mass but not in chemical properties

The percentage of soil components of the soil is: inorganic components

-soil particles

-mineral elements = 45%

-water or moisture= 25%

- air= 25%

organic components

-humus or decayed organic matter= 5%

-living organisms

4 0

3 years ago

A sample of 1.000 g of a compound containing carbon and hydrogen reacts with oxygen at elevated temperature to yield 0.692 g H₂O

ollegr [7]

Answer :

(a) 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) There is no other element present in the compound.

Explanation :

(a) Now we have to determine the masses of C and H in the sample.

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of Carbon and hydrogen respectively.

We are given:

Mass of $CO_2=3.381g$

Mass of $H_2O=0.692g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 3.381 g of carbon dioxide, $\frac{12}{44}\times 3.381=0.922g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.692 g of water, $\frac{2}{18}\times 0.692=0.0769g$ of hydrogen will be contained.

Thus, 1.000 g of compound containing carbon and hydrogen is, 0.922 g and 0.0769 g respectively.

(b) Now we have to determine the compound contain any other elements or not.

Mass carbon + Mass of hydrogen = 0.922 g + 0.0769 g = 0.999 g ≈ 1 g

This means that there is no other element present in the compound.

3 0

3 years ago

Need this ASAP PLEASE

ss7ja [257]

SOLVENT- A substance (usually a liquid) capable of dissolving one or more pure substances. SOLUTE- Solid, liquid or gas that is dissolved in a solvent. SOLUTION- A homogeneous (looks the same throughout) mixture of a solvent and one or more solutes. AQUEOUS SOLUTION- Solution in which water is the solvent.
Answer

it raises the boiling point substance is dissolved in water

7 0

3 years ago