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2 years ago

8

Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and allowed to reach equilibrium described by the equation N2O4(g) 2N

O2(g). If at equilibrium the N2O4 is 28.0% dissociated, what is the value of the equilibrium constant, Kc, for the reaction under these conditions? a. 0.44 b. 2.3 c. 0.11 d. 0.78 e. 0.31

1 answer:

Amanda [17]2 years ago

8 0

Answer : The correct option is, (a) 0.44

Explanation :

First we have to calculate the concentration of $N_2O_4$ .

$\text{Concentration of }N_2O_4=\frac{\text{Moles of }N_2O_4}{\text{Volume of solution}}$

$\text{Concentration of }N_2O_4=\frac{1.0moles}{1.0L}=1.0M$

Now we have to calculate the dissociated concentration of $N_2O_4$ .

The balanced equilibrium reaction is,

$N_2O_4(g)\rightleftharpoons 2NO_2(aq)$

Initial conc. 1.0 M 0

At eqm. conc. (1.0-x) M (2x) M

As we are given,

The percent of dissociation of $N_2O_4$ = $\alpha$ = 28.0 %

So, the dissociate concentration of $N_2O_4$ = $C\alpha=1.0M\times \frac{28.0}{100}=0.28M$

The value of x = $C\alpha$ = 0.28 M

Now we have to calculate the concentration of $N_2O_4\text{ and }NO_2$ at equilibrium.

Concentration of $N_2O_4$ = 1.0 - x = 1.0 - 0.28 = 0.72 M

Concentration of $NO_2$ = 2x = 2 × 0.28 = 0.56 M

Now we have to calculate the equilibrium constant for the reaction.

The expression of equilibrium constant for the reaction will be:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

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

$K_c=\frac{(0.56)^2}{0.72}=0.44$

Therefore, the equilibrium constant $K_c$ for the reaction is, 0.44

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If the composition of the reaction mixture at 400 k is [brcl] = 0.00415 m, [br2] = 0.00366 m, and [cl2] = 0.000672 m, what is th

Scorpion4ik [409]

Q is unlike K value it describes the reaction that is not at equilibrium.
by considering this reaction:
aA+ bB⇄ cC
and our reaction is:
Br2 + Cl2 ⇄ 2 BrCl

According to Q low:
Q= concentration of products/concentration of reactants
but this equation in the gaseous or aqueous states only.
∴ Q = [BrCl]^2 / [Br2] [Cl2]

and we have [Br2] = 0.00366 m [Cl2]= 0.000672 m [BrCl] = 0.00415 m

by substitution:
= [0.00415]^2 / ( [0.00366] * [0.000672])
∴ Q = 7

7 0

2 years ago

2CH4(g)⟶C2H4(g)+2H2(g)

Rasek [7]

Answer : The enthalpy change for the reaction is, 201.9 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 balanced reaction of $CH_4$ will be,

$2CH_4(g)\rightarrow C_2H_4(g)+2H_2(g)$ $\Delta H^o=?$

The intermediate balanced chemical reaction will be,

(1) $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ $\Delta H_1=-890.3kJ$

(2) $C_2H_4(g)+H_2(g)\rightarrow C_2H_6(g)$ $\Delta H_2=-136.3kJ$

(3) $2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_3=-571.6kJ$

(4) $2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(l)$ $\Delta H_4=-3120.8kJ$

Now we will multiply the reaction 1 by 2, revere the reaction 2, reverse and half the reaction 3 and 4 then adding all the equations, we get :

(1) $2CH_4(g)+4O_2(g)\rightarrow 2CO_2(g)+4H_2O(l)$ $\Delta H_1=2\times (-890.3kJ)=-1780.6kJ$

(2) $C_2H_6(g)\rightarrow C_2H_4(g)+H_2(g)$ $\Delta H_2=-(-136.3kJ)=136.3kJ$

(3) $H_2O(l)\rightarrow H_2(g)+\frac{1}{2}O_2(g)$ $\Delta H_3=-\frac{1}{2}\times (-571.6kJ)=285.8kJ$

(4) $2CO_2(g)+3H_2O(l)\rightarrow C_2H_6(g)+\frac{7}{2}O_2(g)$ $\Delta H_4=-\frac{1}{2}\times (-3120.8kJ)=1560.4kJ$

The expression for enthalpy of the reaction will be,

$\Delta H^o=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4$

$\Delta H=(-1780.6kJ)+(136.3kJ)+(285.8kJ)+(1560.4kJ)$

$\Delta H=201.9kJ$

Therefore, the enthalpy change for the reaction is, 201.9 kJ

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2 years ago

What quantum numbers specify these subshells? 2S, 6P, and 3D. (The answer is n= and L=)

statuscvo [17]

N = 1
l = from 0 to (n-1)
ml = -1... + 1
ms = 1/2 or -1/2

eg = 2s
n = 2, m = 0, n = 0
s = 1/2, -1/2

hope this help

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2 years ago

According to the following reaction, how many grams of chloric acid (HClO3) are produced in the complete reaction of 31.6 grams

gogolik [260]

Answer:

$m_{HClO_3}=12.7gHClO_3$

Explanation:

Hello,

Considering the reaction:

$3Cl_2(g)+3H_2O(l)-->5HCl+HClO_3$

The molar masses of chlorine and chloric acid are:

$M_{Cl_2}=35.45*2=70.9g/mol\\M_{HClO_3}=1+35.45+16*3=84.45g/mol$

Now, we develop the stoichiometric relationship to find the mass of chloric acid, considering the molar ratio 3:1 between chlorine and chloric acid, as follows:

$m_{HClO_3}=31.6gCl_2*\frac{1molCl_2}{70.9gCl_2} *\frac{1molHClO_3}{3mol Cl_2} *\frac{85.45g HClO_3}{1mol HClO_3} \\m_{HClO_3}=12.7gHClO_3$

Best regards.

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2 years ago

The density of silver metal bar is 10.5 g/cm3 and it occupies a volume space of 0.5l. What is its mass in kilograms (kg).

mojhsa [17]

Density is equal to the ratio of mass to the volume.

The mathematical expression is given as:

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

Density of silver metal bar= $10.5 g/cm^{3}$

Convert $g/cm^{3}$ into g/L

$1 cm^{3}$ = 0.001 L

Thus, density = $\frac{10.5}{10^{-3}} g/L$

= $10.5\times 10^{3}g/L$

Volume = 0.5 L

Put the values,

$10.5\times 10^{3}g/L= \frac{mass}{0.5 L}$

$m= 10.5\times 10^{3}g/L\times 0.5 L$

= $5.25 \times 10^{3}g$

Now, convert gram into kg

1 g = 0.001 kg

Therefore, mass in kg= $\frac{5.25 \times 10^{3}}{10^{-3}}$

= 5.25 kg

Thus, mass of silver metal bar in kg=5.25 kg

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2 years ago