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2 years ago

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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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In the manufacture of paper, logs are cut into small chips, which are stirred into an alkaline solution that dissolves several o

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Answer:

The estimated feed rate of logs is 14.3 logs/min.

Explanation:

The product of the process is 2000 tons/day of dry wood pulp, of 85 wt% of cellulose. That represents (2000*0.85)=1700 tons/day of cellulose.

That cellulose has to be feed by the wood chips, which had 47 wt% of cellulose in its composition. That means you need (1700/0.47)=3617 tons/day of wood chips to provide all that cellulose.

Th entering flow is wood chips with 45 wt% of water. This solution has an specific gravity of 0.640.

To know the specific gravity of the wood chips we have to write a volume balance. We also know that Mw=0.45*M and Mc=0.55*M.

$V=V_c+V_w\\\\M/\rho=M_c/\rho_c+Mw/\rho_w\\\\M/\rho=0.55*M/\rho_c+0.45*M/\rho_w\\\\1/\rho=0.55/\rho_c +0.45/\rho_w\\\\0.55/\rho_c=1/\rho-0.45/\rho_w\\\\0.55/\rho_c=1/(0.64*\rho_w)-0.45/\rho_w=(1/\rho_w)*(\frac{1}{0.64}-\frac{0.45}{1} )\\\\0.55/\rho_c=1.1125/\rho_w\\\\\rho_c=\frac{0.55}{1.1125}*\rho_w= 0.494*\rho_w$

The specific gravity of the wood chips is 0.494.

The average volume of a log is

$V_l=(\pi*D^{2} /4)*L=(3.1416*\frac{8^{2} \, in^{2} }{4} )*9ft*(\frac{12 in}{1ft})= 21714 in^{3}=12.57 ft^{3}$

The weight of one log is

$M=\rho*V=0.494*\rho_w*12.57 ft^{3}\\\\M=0.494*62.4\frac{lbm}{ft^{3} }*12.57ft^{3}\\\\M=387.5lbm$

To provide 3617 ton/day of wood chips, we need

$n=\frac{supply}{M_{log}}=\frac{3617 tons/day}{387.5 lbm}*\frac{2204lbm}{1ton}\\\\n=20573 logs/day=14.3 logs/min$

The feed rate of logs is 14.3 logs/min.

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At high temperature, 2.00 mol of HBr was placed in a 4.00 L container where it decomposed in the reaction: 2HBr(g) H2(g) Br2(g)

viktelen [127]

Answer: $K_c$ for this reaction at this temperature is 0.029

Explanation:

Moles of $HBr$ = 2.00 mole

Volume of solution = 4.00 L

Initial concentration of $HBr=\frac{moles}{Volume}=\frac{2.00}{4.00L}=0.500M$

The given balanced equilibrium reaction is,

$2HBr(g)\rightleftharpoons H_2(g)+Br_2(g)$

Initial conc. 0.500 M 0 M 0 M

At eqm. conc. (0.500-2x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[H_2\times [Br_2]}{[HBr]^2}$

Equilibrium concentration of $[Br_2]$ = x = 0.0955 M

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

$K_c=\frac{0.0955\times 0.0955}{0.500-2\times 0.0955}$

$K_c=0.029$

Thus $K_c$ for this reaction at this temperature is 0.029

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A 0.3870-g sample of a compound known to contain only carbon, hydrogen, and oxygen was burned in oxygen to yield 0.7191Â g of co

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The chemical formula for the compound can be written as,

CxHyOz

where x is the number of C atoms, y is the number of H atoms, and z is the number of O atoms. The combustion reaction for this compound is,

CxHyOz + O2 --> CO2 + H2O

number of moles of C:
(0.7191 g)(1 mol CO2/44 g of CO2) = 0.0163 mol CO2
This signifies that 0.0163 mole of C and the mass of carbon in the compound,
(0.0163 mols C)(12 g C/ 1 mol C) = 0.196 g C

number of moles H:
(0.1472 g H2O)(1 mol H2O/18 g H2O) = 0.00818 mol H2O

This signifies that there are 0.01635 atoms of H in the compound.
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0.3870 - (0.196 g C + 0.01635 g H) = 0.1746 g

Moles O in the compound = (0.1746 g O)(1 mol O/16 g O) = 0.0109 mols O

The formula of the compound is,
C0.0163H0.01635O0.0109

Dividing the numbers by the least number,
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The empirical formula of the compound is therefore,
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Under standard conditions, a given reaction is endergonic (i.e., ΔG >0). Which of the following can render this reaction favo

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Answer:

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