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Lapatulllka [165]
3 years ago
5

Butane (C4H10(g), deltaHf =-125.6 kJ/mol reacts with oxygen to produce carbon dioxide (CO2 delta Hf =-393.5 kJ/mol)

Chemistry
1 answer:
schepotkina [342]3 years ago
7 0

The given question is incomplete. The complete question is:

Butane (C_4H_{10}(g), Hf = –125.6 kJ/mol) reacts with oxygen to produce carbon dioxide (CO_2 , Hf = –393.5 kJ/mol ) and water (H_2O, Hf = –241.82 kJ/mol) according to the equation below. What is the enthalpy of combustion (per mole) of C_4H_{10} (g)?

Answer: -2657.5 kJ

Explanation:

The balanced chemical reaction is,

C_4H_{10}(g)+\frac{13}{2}O_2(g)\rightarrow 4CO_2(g)+5H_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_4H_{10}}\times \Delta H_{C_4H_{10})]

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

\Delta H=[(4\times -393.5)+(5\times -241.82)]-[(\frac{13}{2}\times 0)+(1\times -125.6)]

Delta H=-2657.5kJ

Therefore, the enthalpy of combustion per mole of butane is -2657.5 kJ

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∴ Kp = 3.7*10^-3 (0.0821* 1000)^2= 24.939



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