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Effectus [21]
3 years ago
7

The standard heats of formation for CO2(g), C2H6(g), and H2O(l) are -394.0 kJ/mol, -84.00 kJ/mol, and -286.0 kJ, respectively. W

hat is the standard heat of reaction for the following process? C2H6(g) + 7/2 O2(g) --> 2CO2(g) + 3H2O(l)
Chemistry
1 answer:
Ivanshal [37]3 years ago
6 0

Answer:

ΔH°r = -1562 kJ

Explanation:

Let's consider the following combustion.

C₂H₆(g) + 7/2 O₂(g) ⇒ 2 CO₂(g) + 3 H₂O(l)

We can calculate the standard heat of reaction (ΔH°r) using the following expression:

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

ni are the moles of reactants and products

ΔH°f(i) are the standard heats of formation of reactants and products

The standard heat of formation of simple substances in their most stable state is zero. That means that ΔH°f(O₂(g)) = 0

ΔH°r = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

ΔH°r = [2 mol × ΔH°f(CO₂) + 3 mol × ΔH°f(H₂O)] - [1 mol × ΔH°f(C₂H₆) + 7/2 mol × ΔH°f(O₂)]

ΔH°r = [2 mol × (-394.0 kJ/mol) + 3 mol × (-286.0 kJ/mol)] - [1 mol × (-84.00 kJ/mol) + 7/2 mol × 0]

ΔH°r = -1562 kJ

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Balanced chemical reaction:

C₃H₈(g) + 3H₂O(g) → 3CO(g) + 7H₂(g).

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Consider the reaction for the decomposition of hydrogen disulfide: 2H2S(g)⇌2H2(g)+S2(g), Kc = 1.67×10−7 at 800∘C A 0.500 L react
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Answer:

Molar concentration of S₂ is 1.77×10⁻⁶M

Explanation:

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Concentrations in equilibirum are:

[H₂S] : 0,163/0.500L - X

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Replacing the concentrations and the equilibrium value in (1):

K = \frac{[X][0.1+X]^2}{[0326-X]^2}

1.67x10⁻⁷ = X (X² + 0.2X + 0.01) / (X² -0.652X + 0.106)

1.67x10⁻⁷X² - 1.09x10⁻⁷X + 1.77x10⁻⁸ = X³ + 0.2X² + 0.01X

0 =  X³ + 0.2X² + 0.01X - 1.77x10⁻⁸

Solving for X:

X = 1.77×10⁻⁶

As [S₂] = X, <em>molar concentration of S₂ is 1.77×10⁻⁶M</em>

I hope it helps!

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