What should be the Δ f H 0 ( CH 4 , g ) in kJ / mol . | TestHub

Passage / Comprehension

The enthalpy of combustion of a given compound is defined as the change in enthalpy when 1 mole of this compound combines with the required amount of O₂ to give products in their stable forms. The enthalpy of combustion is usually measured by placing a known mass of the compound in a closed steel container (known as a bomb calorimeter).

$Δ H_{co mb}^{0} (C H_{4,} g) = - 900 k J$ /mole

$Δ H_{co mb}^{0} (C, g r a p hi t e) = - 400 k J$ /mole

$Δ H_{co mb}^{0} (H_{2}, g) = - 300 k J$ /mole

$Δ H_{a t o mi z a t i o n}^{0} (C, g r a p hi t e) = 700 k J$ /mole

and $Δ H_{f}^{0} (H, g)$ $= 220 kJ/mol$

What should be the $Δ_{f} H^{0} (CH_{4}, g)$ in $kJ / mol$ .

Answer:

-100.00

Solution:

Question Explanation: Calculate the standard enthalpy of formation of methane $(CH_{4})$ . Concept: Hess's Law of Constant Heat Summation.

Solution:

The enthalpy of formation is found by subtracting the enthalpy of combustion of the compound from the sum of combustion enthalpies of its elements.

C(graphite) + 2H₂(g) → CH₄(g)

$Δ_{f} H_{0} (C H_{4}) =$ $Δ_{c} H_{0} (C, g r a p hi t e) + 2$ [ $Δ_{c} H_{0} (H_{2}, g)$ - $Δ_{c} H_{0} (C H_{4}, g)$

$Δ_{f} H_{0} = - 400 + 2 (- 300) - (- 900)$

$Δ_{f} H_{0}$ $= - 400 - 600 + 900 = - 100 k J$ /mole

Chemistry - Thermochemistry Question with Solution | TestHub

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