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First Law of Thermodynamics

Refer to the ...

Question

Refer to the following potential energy diagram and the choices above,
What is the $△ H$ of the reaction to form $C O$ from $C + O_{2}$

- 94.0 k c a l

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- 26.4 k c a l

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- 67.6 k c a l

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C (s) + \frac{1}{2} O_{2} (g)

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C O_{2} (g)

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Solution

The correct option is B $- 26.4 k c a l$
$C (s) + \frac{1}{2} O_{2} \to C O (g)$
$Δ H_{f} (C O) = Δ H_{f} (C (s)) + Δ H_{f} (O_{2} (g))$
$Δ H_{f} (C O) = - 26.4 k c a l$
$Δ H_{f} (O_{2} (g)) = 0$ because enthalpy of formation for naturally existing elements is zero.

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Similar questions

△ H

C O_{2}

C + O_{2}

C (s) + O_{2} (g) \to C O_{2} (g), Δ H = + 94.0 K c a l .

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g), Δ H = - 67.7 K c a l .

^{- 1}

C (s) + \frac{1}{2} O_{2} (g) \to C O (g)

△ H

C O_{2}

C O + O_{2}

C (s) + \frac{1}{2} O_{2} (g) \to C O (g); Δ H_{1} = a

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ H_{2} = b

C (s) + C O_{2} (g) \to 2 C O (g); Δ H_{3} = c

C (S) + O_{2} (g) \to C O_{2} (g)

Δ H = - 94.0 K c a l

H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l)

Δ H = - 68.0 K c a l

C H_{3} C O O H (l) + 2 O_{2} (g) \to 2 C O_{2} (g) + 2 H_{2} O (l)

Δ H = - 210.0 K c a l

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