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3 years ago

Give the ΔH value for the combustion of butane as shown in the reaction 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(g)+5315 kJ.

Chemistry

2 answers:

777dan777 [17]3 years ago

7 0

Answer:

ΔH = - 5315 kJ.

Explanation:

The given chemical reaction is as follows -

2C₄H₁₀ (g) + 13 O₂ (g) → 8 CO₂ (g) + 10 H₂O (g) + 5315 kJ

In the above equation , the amount of energy i.e. 5315 kJ is released , i.e. it is in the product side , hence , the reaction is an example of an exothermic reaction .

Hence ,

The value of the change in enthalphy , i.e. , the enthalpy of product minus the enthalpy of the product .

Therefore ,

The value of the change in enthalphy = - ve .

Hence ,

ΔH = - 5315 kJ.

antiseptic1488 [7]3 years ago

5 0

Answer:

ΔH° = -3790 kJ

Explanation:

Let's consider the combustion of butane.

2 C₄H₁₀(g) +13 O₂(g) → 8 CO₂(g) + 10 H₂O(g)

We can find the standard enthalpy of this combustion reaction (ΔH°) using the following expression.

ΔH° = ∑np × ΔH°f(p) - ∑nr × ΔH°f(r)

where,

n: mole

ΔH°f(): standard enthalpy of formation

p: product

r: reactant

ΔH° = 8 mol × ΔH°f(CO₂(g)) + 10 mol × ΔH°f(H₂O(g)) - 2 mol × ΔH°f(C₄H₁₀(g)) - 13 mol × ΔH°f(O₂(g))

ΔH° = 8 mol × (-393.5 kJ/mol) + 10 mol × (-241.8 kJ/mol) - 2 mol × (-888.0 kJ/mol) - 13 mol × 0

ΔH° = -3790 kJ

Given ΔH° < 0, the reaction is exothermic.

2 C₄H₁₀(g) +13 O₂(g) → 8 CO₂(g) + 10 H₂O(g) + 3790 kJ

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2 years ago

Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163

statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

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3 years ago

Calculate the de broglie wavelength of a subatomic particle that is moving at 351 km/s if its mass is 9.11 Ã 10â31 kg. Î» = hmuh

gregori [183]

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Answer:

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Explanation:

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2 years ago

What is the (OH-) in a solution with a pOH of 6.48

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3 years ago