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Alekssandra [29.7K]
3 years ago
12

Consider the following numbered processes: 1. A → 2B 2. B → C + D 3. E → 2D ΔH for the process A → 2C + E is

Chemistry
1 answer:
aivan3 [116]3 years ago
6 0

Answer:

ΔH = ΔH₁ + ΔH₂ - ΔH₃

Explanation:

Given that:

1. A → 2B

2. B → C + D

3. E → 2D

Assuming from the corresponding ΔH for process 1, 2 and 3 are ΔH₁, ΔH₂, ΔH₃ respectively.

To estimate the ΔH for the process A → 2C + E

We multiply 2 with equation 2 where (B → C + D)

2B → 2C + 2D ⇒ 2ΔH₂

Also, let's switch equation (3), such that we have,

2D → E -ΔH₃

The summation of all the equation result into :

A → 2C + E

where; ΔH = ΔH₁ + ΔH₂ - ΔH₃

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(a) What is the total volume (in L) of gaseous products, measured at 350°C and 735 torr, when an automobile engine burns 100. g
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Answer:

Part A

 The volume of the gaseous product  is  V = 787L

Part B

The volume of the the engine’s gaseous exhaust is  V_e = 2178 \ L

Explanation:

Part A

From the question we are told that

    The temperature is  T = 350^oC = 350 +273 =623K

     The pressure is  P = 735 \ torr = \frac{735}{760} =  0.967\ atm

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The chemical equation for this combustion is

               2 C_8 H_{18}_{(l)} + 25O_2_{(l)} ----> 16CO_2_{(g)} + 18 H_2 O_{(g)}

 The number of moles of  C_8 H_{18} that reacted is mathematically represented as

               n = \frac{mass \ of \  C_8H_{18}  }{Molar \  mass \ of  C_8H_{18} }

The molar mass of  C_8 H_{18} is constant value which is

                  M = 114.23 \ g/mole  

So          n = \frac{100  }{114.23} }

             n = 0.8754 \ moles

The gaseous product in the reaction is CO_2_{(g)} and water vapour

Now from the reaction

    2 moles of C_8 H_{18}  will react with 25 moles of O_2 to give (16 + 18) moles of CO_2_{(g)} and  H_2 O_{(g)}

So

    1 mole of C_8 H_{18} will  react with 12.5 moles of  O_2 to give 17 moles of CO_2_{(g)} and  H_2 O_{(g)}

This implies that

    0.8754 moles of C_8 H_{18} will react with (12.5 * 0.8754 ) moles of O_2 to give  (17 * 0.8754) of CO_2_{(g)} and  H_2 O_{(g)}

So the no of moles of gaseous product is

         N_g = 17 * 0.8754

         N_g = 14.88 \ moles

From the ideal gas law

       PV = N_gRT

making V the subject

        V = \frac{N_gRT}{P}

Where R is the gas constant with a value R = 0.08206 \  L\cdot atm /K \cdot mole

Substituting values

          V = \frac{14.88* 0.08206 *623}{0.967}

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Part B

From the reaction the number of moles of oxygen that reacted is

         N_o = 0.8754 * 12.5

         N_o = 10.94 \ moles

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      V_o  = \frac{10.94 * 0.08206 *623}{0.967}

      V_o  = 579 \ L

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         V_e = V_o * \frac{0.79}{0.21}

Substituting values

       V_e = 579 * \frac{0.79}{0.21}

       V_e = 2178 \ L

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