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timurjin [86]
3 years ago
9

Find the molar enthalpy of formation for paraffin wax (C2H526)) given the following reaction

Chemistry
1 answer:
pogonyaev3 years ago
3 0

Answer: The molar enthalpy of formation for paraffin wax  is  -2460.5 kJ

Explanation:

The balanced chemical reaction is,

C_{25}H_{52}(g)+38O_2(g)\rightarrow 25CO_2(g)+26H_2O(g)

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_{25}H_{52}}\times \Delta H_{C_{25}H_{52}})]

where,

n = number of moles

\Delta H_{O_2}=0 (as heat of formation of substances in their standard state is zero

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

-14800=[(25\times -393.5)+(26\times -285.5)]-[(38\times 0)+(1\times \Delta H_{C_{25}H_{52}})]

\Delta H_{C_{25}H_{52}}=-2460.5kJ/mol

Therefore, the molar enthalpy of formation for paraffin wax  is -2460.5 kJ

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Propane is often used to heat homes. The combustion of propane follows the following reaction: C3H8(g) + 5O2(g)  3CO2 (g) + 4H
swat32

Answer:

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

Explanation:

<u>Step 1</u>: Data given

C3H8 + 5O2 -----------> 3CO2 + 4H2O      ΔH° = –2044 kJ

This means every mole C3H8

Every mole of C3H8 produces 2044 kJ of heat when it burns (ΔH° is negative because it's an exothermic reaction)

<u>Step 2: </u>Calculate the number of moles to produce 7563 kJ of heat

1 mol = 2044 kJ

x mol = 7563 kJ

x = 7563/2044 =  3.70 moles

To produce 7563 kJ of heat we have to burn 3.70 moles of C3H8

<u>Step 3: </u>Calculate mass of propane

Mass propane = moles * Molar mass

Mass propane = 3.70 moles * 44.1 g/mol

Mass propane = 163.17 grams

To release 7563 kJ of heat, we need to burn 163.17 grams of propane

7 0
3 years ago
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