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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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For a given chemical reaction, the addition of a catalyst provides a different pathway that(1) decreases the reaction rate and h
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Answer: Option (4) is the correct answer.

Explanation:

Activation energy is the minimum amount of energy required by the reactants to participate in the chemical reaction.

Therefore, reactant molecules whose energy is less than the activation energy are not able to participate in the reaction. Hence, a catalyst lowers the activation energy so that molecules with less energy can also participate in the reaction.

Thus, a catalyst is a specie that increases the reaction rate and has a lower activation energy.

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2 years ago
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nydimaria [60]
First, let us write the reaction. We know that the reactants are Aluminum and Sulfuric acid, and one of the products is hydrogen. That means the reaction is a single replacement reaction as shown below:

2 Al + 3 H₂SO₄ = 3 H₂ + Al₂(SO₄)₃

Next, let us determine which of the reactants is limiting:

1.80 g Al (1 mol/26.98 g)(3 mol H₂SO₄/2 mol Al)(98 g H₂SO₄/mol) = 9.807 g

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2 years ago
Energy and Enthalpy Changes, Heat and Work -- Monatomic Ideal Gas dynamically generated plot 2.00-mol of a monatomic ideal gas g
larisa86 [58]

(a) The heat generated in the process is 28 kJ.

(b) The work done in the process is determined as -28 kJ.

(c) The change in the internal energy is 0.

<h3>Heat of the isothermal compression </h3>

The heat generated in the process is negative done in the process.

W = -PΔV

W = -P(V₂ - V₁)

<h3>From A to B</h3>

W = -P(VB - VA)

W = -11(7 - 12.5)

W = 60.5 L.atm = 60.5 x 101.325 J/L.atm = 6,130.16 J

<h3>From C to D</h3>

W = -25(20.5 - 7)

W = -337.5 L.atm = -34,197.18 J

Total work , w = -34,197.18 J +  6,130.16 J = -28 kJ

q = - w

q = 28 kJ

<h3>Change in internal energy</h3>

ΔE = q + w

ΔE = 28 kJ - 28 kJ = 0

Learn more about change in internal energy here: brainly.com/question/17136958

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mestny [16]
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