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-Dominant- [34]
3 years ago
9

For the reaction CO+2H2=CH3OH at 700 K, equilibrium concentrations are [H2]=0.072 M, [CO]= 0.020M, and [CH3OH]= 0.030 M. Calcula

te K
Chemistry
1 answer:
bekas [8.4K]3 years ago
3 0

The balanced equation for the reaction is

CO(g) + 2H₂(g) ⇄ CH₃OH(g)

 

The given concentrations are at equilibrium state. Hence we can use them directly in calculation with the expression for the equilibrium constant, k. expression for k can be written as

   k = [CH₃OH(g)] / [CO(g)] [H₂<span>(g) ]²

</span>[H₂<span>]=0.072 M
[CO]= 0.020M
[CH</span>₃OH]= 0.030 M

 

From substitution,

   k = 0.030 M / 0.020 M x (0.072 M)²

   k = 289.35 M⁻²

<span>
Hence, equilibrium constant for the given reaction at 700 K is 289.35 M</span>⁻².

<span> </span>

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A solution is prepared by mixing 250 mL of 1.00 M CH3COOH with 500 mL of 1.00 M NaCH3COO. What is the pH of this solution? (Ka f
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Answer:

A solution is prepared by mixing 250 mL of 1.00 M

CH3COOH with 500 mL of 1.00 M NaCH3COO.

What is the pH of this solution?

(Ka for CH3COOH = 1.8 × 10−5 )

Explanation:

This is a case of a neutralization reaction that takes place between acetic acid,     CH 3 COOH ,   a weak acid, and sodium hydroxide,   NaOH , a strong base.

The resulting solution pH, depends if the neutralization is complete or not.  If not, that is, if the acid is not completely neutralized, a buffer solution containing acetic acid will be gotten, and its conjugate base, the acetate anion.

It's important to note that at complete neutralization, the pH of the solution will not equal  7 . Even if the weak acid is neutralized completely, the solution will be left with its conjugate base, this is the reason why the expectations of its pH is to be over  7 .

So, the balanced chemical equation for this reaction is the ionic equation:

CH 3 COOH (aq]  +  OH − (aq]  →  CH 3 COO − (aq]  +  H 2 O (l]

Notice that:  

1  mole of acetic acid will react with:  1  mole of sodium hydroxide, shown here as hydroxide anions,  OH − , to produce   1   mole of acetate anions:

CH 3 COO −

To determine how many moles of each you're adding , the molarities and volumes of the two solutions are used:

     c  =  n /  V    ⇒     n   =   c  ⋅  V

n  acetic   =   0.20 M   ⋅   25.00   ⋅   10  − 3 L   =   0.0050 moles CH3 COOH

and

n  hydroxide   =   0.10 M   ⋅   40.00   ⋅   10 − 3 L   =   0.0040 moles OH −

There are fewer moles of hydroxide anions, so the added base will be completely consumed by the reaction.

As a result, the number of moles of acetic acid that remain in solution is:

    n  acetic remaining   =   0.0050  −   0.0040   =    0.0010 moles

The reaction will also produce  0.0040   moles of acetate anions.

This is, then a buffer and the Henderson-Hasselbalch equation is applied to find its pH :

pH  =  p K a  +  log  ( [ conjugate base ]  / [ weak acid ] )

Use the total volume of the solution to find the new concentrations of the acid and of its conjugate base .

V total  =  V acetic  +  V hydroxide

V total  =  25.00 mL  +  40.00 mL  =  65.00 mL

Thus the concentrations will be :

[ CH 3 COOH ]  =  0.0010 moles  / 65.00  ⋅  10 − 3 L  =  0.015385 M

and

[ CH 3 COO − ]  =  0.0040 moles  / 65  ⋅  10 − 3 L  =  0.061538 M

The    p K a     of acetic acid is equal to    4.75

Thus the pH of the solution will be:

pH   =   4.75  +  log ( 0.061538 M  /    0.015385 M )

pH   =   5.35

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Answer:

The molecular formula = C_2Cl_{4}F_2

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 11.79

Molar mass of C = 12.0107 g/mol

<u>% moles of C = 11.79 / 12.0107 = 0.9816</u>

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Molar mass of Cl = 35.453 g/mol

<u>% moles of Cl = 69.57 / 35.453 = 1.9623</u>

Given that the gaseous chlorofluorocarbon only contains chlorine, flourine and carbon. So,

% of F = 100% - % of C - % of C = 100 - 11.79 - 69.57 = 18.64

Molar mass of F = 18.998 g/mol

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Taking the simplest ratio for C, Cl and F as:

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= 1 : 2 : 1

The empirical formula is = CCl_2F

Also, Given that:

Pressure = 21.3 mm Hg

Also, P (mm Hg) = P (atm) / 760

Pressure = 21.3 / 760 = 0.02803 atm

Temperature = 25 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

Volume = 458 mL  = 0.458 L (1 mL = 0.001 L)

Using ideal gas equation as:

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.02803 atm × 0.458 L = n × 0.0821 L.atm/K.mol × 298.15 K  

⇒n = 0.00052445 moles

Given that :  

Amount  = 0.107 g  

Molar mass = ?

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.00052445= \frac{0.107\ g}{Molar\ mass}

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Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 1×12.0107 + 2×35.453 + 1×18.998 = 101.9147 g/mol

Molar mass = 204.0233 g/mol

So,  

Molecular mass = n × Empirical mass

204.0233 = n × 101.9147

⇒ n = 2

<u>The molecular formula = C_2Cl_{4}F_2</u>

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