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victus00 [196]
2 years ago
13

Ethanol (C2H5OH) melts at –114 °C and boils at 78 °C. The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vapo

rization is 38.56 kJ/mol. The specific heats of solid and liquid ethanol are 0.97 J/g-K and 2.3 J/g-K, respectively. The average specific heat of gaseous ethanol is about 1.80 J/g-K. a. How much heat is required to convert 35.0 g of ethanol at 27 °C to the vapor phase at 120 °C? b. How much heat is required to convert the same amount of ethanol at –120 °C to the vapor phase at 120 °C?
Chemistry
1 answer:
alexandr1967 [171]2 years ago
5 0

Answer:

First question

       Q =  36826 \  J

Second  question  

       Q =  52299.7  \  J

Explanation:

From the question we are told that

     The melting point of Ethanol is  T_m  =  -114 ^oC

      The boiling point of Ethanol is  T_b   =  78^ oC

       The enthalpy of fusion of Ethanol is F  =  5.02 \  kJ / mol = 5.02 *10^{3}\  kJ / mol

        The enthalpy of vaporization  of Ethanol is L  =  38.56  \  kJ / mol = 38.56 *10^{3} \  J / mol

         The specific heat of solid Ethanol is  c_e = 0.97 \  J/ g \cdot K

          The specific heat of liquid  Ethanol is c_l  =  2.3 \  J / g \cdot K

           The mass of the Ethanol given is  m =  35.0 \ g

Considering the first question

           The initial  temperature is T_i  =  27^oC

             The final  temperature is  T_f  =  120^oC

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

       Q_1 =  m  *  c_l *  (T_b - T_i)

=>      Q_1  =  35.0   *  2.3  *  (  78 - 27)

=>      Q_1  =4106 \ J

Genially the number of moles of Ethanol given is mathematically represented as

         n  = \frac{m}{Z}

Here Z  is the molar mass of Ethanol  with value  Z =  46 g/mol

So

         n  = \frac{35}{46 }

=>      n  = 0.7609 \  mol

Generally the heat of vaporization of the Ethanol is mathematically represented as

         Q_2 = n  * L

=>        Q_2  =0.7809   *  38.56  * 10^{3}

=>        Q_2  =29339  \  J

Generally the heat required too raise the Ethanol from  its boiling point to  T_f  is  mathematically represented as

       Q_3 =  m  *  c_l *  (T_f - T_b)

=>     Q_3 =  35   *   2.3 *  (120 - 78 )

=>     Q_3 = 3381 \  J

Generally the total heat required is  

     Q =  Q_1 + Q_2 + Q_3

=>   Q =  4106    + 29339   + 3381

=>   Q =  36826 \  J

Considering the second question

           The initial  temperature is T_i  =  -120^oC

             The final  temperature is  T_f  =  120^oC

Generally the heat required too raise the Ethanol to its melting  point is mathematically represented as

       Q_1 =  m  *  c_e *  (T_m - T_i)

=>      Q_1  =  35.0   *  0.97  *  ( -114 - (- 120) )

=>      Q_1  = 203.7 \ J

Generally the heat of fusion  of the Ethanol is mathematically represented as

                 Q_2 = n  * F

=>        Q_2  =0.7809   *  5.02 *10^{3}

=>        Q_2  =3920  \  J

Generally the heat required too raise the Ethanol to its boiling point is mathematically represented as

       Q_3 =  m  *  c_l *  (T_b - T_m)

=>      Q_3  =  35.0   *  2.3  *  ( 78 - (- 114) )

=>      Q_3  =15456  \ J

Generally the heat of vaporization of the Ethanol is mathematically represented as

         Q_4 = n  * L

=>        Q_4  =0.7809   *  38.56  * 10^{3}

=>        Q_4  =29339  \  J

Generally the heat required too raise the Ethanol from  its boiling point to  T_f  is  mathematically represented as

       Q_5 =  m  *  c_l *  (T_f - T_b)

=>     Q_5 =  35   *   2.3 *  (120 - 78 )

=>     Q_5 = 3381 \  J

Generally the total heat required is  

     Q =  Q_1 + Q_2 + Q_3+Q_4 + Q_5

=>   Q =  203.7    + 3920   + 15456 +29339+3381

=>   Q =  52299.7  \  J

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Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC

And the moles of hydrogen due to the produced 1.50 grams of water:

n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}  =0.166molH

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO

Thus, the subscripts in the empirical formula are:

C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol

Which are equal to the moles of the acid:

n_{acid}=0.0279mol

And the molar mass:

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C_7H_6O_2

Best regards!

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