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Arte-miy333 [17]

3 years ago

5

The combustion of 0.0222g of isooctane vapor, C8H18(g), at constant pressure raises the temperature of a calorimeter 0.400 degre

es C. The heat capacity of the calorimeter and water combined is 2.48kJ/degrees C. Find the molar heat of combustion of gaseous isooctane.

1 answer:

saul85 [17]3 years ago

4 0

Answer:

ΔH = 5102.88 KJ/mol

Explanation:

Qp = ΔH = CpΔT... constant pressure

∴ Cp = 2.48 KJ/°C

∴ ΔT = 0.400°C

∴ Mw C8H18(g) = 114.22 g/mol

∴ mol C8H18(g) = (0.0222 g)×(mol C8H18/114.22 g) = 1.944 E-4 mol C8H18

⇒ ΔH = (2.48 KJ/°C)×(0.400°C) = 0.992 KJ

⇒ ΔH = 0.992 KJ/ 1.944 E-4 mol C8H18

⇒ ΔH = 5102.88 KJ/mol

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According to the Arrhenius equation,

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The expression used with catalyst and without catalyst is,

$\frac{K_2}{K_1}=\frac{A\times e^{\frac{-Ea_2}{RT}}}{A\times e^{\frac{-Ea_1}{RT}}}$

$\frac{K_2}{K_1}=e^{\frac{Ea_1-Ea_2}{RT}}$

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$K_2$ = rate constant reaction -1

$K_1$ = rate constant reaction -2

Activation energy of the reaction 1 , $Ea_1$ = 14.0 kJ/mol = 14,000 J

Activation energy of the reaction 2, $Ea_1$ = 11.9 kJ/mol = 11,900 J

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T = temperature = $25^oC=273+25=298 K$

Now put all the given values in this formula, we get

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4moles of Fe produced 2moles of Fe2O3.

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Now we need to find the mass present in 0.09mol of Fe2O3. This can be achieved by doing the following:

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$[C6H5COO^{-}] = \frac{[1.1\times 10^{-5}][1.1\times 10^{-5}]}{1.6\times 10^{-10}}$

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So, Concentration of NaC6H5COO would also be 0.76 M and volume is given to us 0.50 L , now moles can we calculated as : Moles = M X L

Moles of NaC6H5COO would be = $0.76(\frac{mol}{L}) \times (0.50L)$

Moles of NaC6H5COO (sodium benzoate) = 0.38 mol

8 0

3 years ago

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