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3 years ago

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The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate

the value of pKa for

1 answer:

Tems11 [23]3 years ago

6 0

Answer:

6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

pKa =.?

Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

Finally, we shall determine the pka for the reaction as follow:

Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

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Alenkasestr [34]

Answer : The value of $\Delta H_{rxn}$ for the reaction is, -390.3 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The main chemical reaction is:

$CH_4(g)+4Cl_2(g)\rightarrow CCl_4(g)+4HCl(g)$ $\Delta H_{rxn}=?$

The intermediate balanced chemical reaction are:

(1) $C(s)+2H_2(g)\rightarrow CH_4(g)$ $\Delta H_1=-74.6kJ$

(2) $C(s)+2Cl_2(g)\rightarrow CCl_4(g)$ $\Delta H_2=-95.7kJ$

(3) $H_2(g)+Cl_2(g)\rightarrow 2HCl(g)$ $\Delta H_2=-184.6kJ$

Now we are reversing reaction 1, multiplying reaction 3 by 2 and then adding all the equations, we get :

(1) $CH_4(g)\rightarrow C(s)+2H_2(g)$ $\Delta H_1=74.6kJ$

(2) $C(s)+2Cl_2(g)\rightarrow CCl_4(g)$ $\Delta H_2=-95.7kJ$

(3) $2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)$ $\Delta H_2=2\times (-184.6kJ)=-369.2kJ$

The expression for enthalpy change for the reaction will be,

$\Delta H_{rxn}=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H_{rxn}=(74.6kJ)+(-95.7kJ)+(-369.2kJ)$

$\Delta H_{rxn}=-390.3kJ$

Therefore, the value of $\Delta H_{rxn}$ for the reaction is, -390.3 kJ

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3 years ago

Given the following equation: 2 C4H10 13 O2 > 8 CO2 10 H20 + How many grams of CO2 are produced if 12.4 grams of C4H10 reacts

ahrayia [7]

Answer : The mass of $CO_2$ produced will be, 37.488 grams.

Explanation : Given,

Mass of $C_4H_{10}$ = 12.4 g

Mass of $O_2$ = 56.9 g

Molar mass of $C_4H_{10}$ = 58 g/mole

Molar mass of $O_2$ = 32 g/mole

Molar mass of $CO_2$ = 44 g/mole

First we have to calculate the moles of $C_4H_{10}$ and $O_2$ .

$\text{Moles of }C_4H_{10}=\frac{\text{Mass of }C_4H_{10}}{\text{Molar mass of }C_4H_{10}}=\frac{12.4g}{58g/mole}=0.213moles$

$\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{56.9g}{32g/mole}=1.778moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$2C_4H_{10}+13O_2\rightarrow 8CO_2+10H_2O$

From the balanced reaction we conclude that

As, 2 moles of $C_4H_{10}$ react with 13 mole of $O_2$

So, 0.213 moles of $C_4H_{10}$ react with $\frac{13}{2}\times 0.213=1.385$ moles of $O_2$

From this we conclude that, $O_2$ is an excess reagent because the given moles are greater than the required moles and $C_4H_{10}$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $CO_2$ .

As, 2 moles of $C_4H_{10}$ react to give 8 moles of $CO_2$

So, 0.213 moles of $C_4H_{10}$ react to give $\frac{8}{2}\times 0.213=0.852$ moles of $CO_2$

Now we have to calculate the mass of $CO_2$ .

$\text{Mass of }CO_2=\text{Moles of }CO_2\times \text{Molar mass of }CO_2$

$\text{Mass of }CO_2=(0.852mole)\times (44g/mole)=37.488g$

Therefore, the mass of $CO_2$ produced will be, 37.488 grams.

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poizon [28]

$\quad \quad\quad\huge \tt{ANSWER}$

Hydrocarbons are organic compounds which made up of only carbon and hydrogen.

$\pink{ \star}$ Carbon - ( C )

$\pink{ \star}$ Hydrogen - ( H )

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

$\red{✿}$ Type of Hydrocarbon :

<em>Alkanes</em>

<em>Alkenes</em>

<em>Alkynes</em>

$\purple{\mathfrak{llXecretll}}$

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1 year ago

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vova2212 [387]

Answer:

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