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slega [8]
3 years ago
11

Find ΔHrxn for the following reaction: 2PbS(s)+3O2(g)→2PbO(s)+2SO2(g)

Chemistry
1 answer:
ch4aika [34]3 years ago
5 0

Answer:

ΔH°rxn = -827.5 kJ

Explanation:

Let's consider the following balanced equation.

2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)

We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]

ΔH°rxn = -827.5 kJ

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3 years ago
I have 50.00 mL of 0.100 M ethyl amine (C2H5NH2). I gradually add a solution of 0.025 M nitric acid (HNO3) to the ethyl amine so
sergeinik [125]

Answer:

4.00 is the pH of the mixture

Explanation:

The ethyl amine reacts with HNO3 as follows:

C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻

To solve this question we need to find the moles of ethyl amine and the moles of HNO3:

<em>Moles C2H5NH2:</em>

0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine

<em>Moles HNO3:</em>

0.201L * (0.025mol/L) = 0.005025 moles HNO3

That means HNO3 is in excess. The moles in excess are:

0.005025 moles HNO3 - 0.00500 moles ethyl amine =

2.5x10⁻⁵ moles HNO₃

In 50 + 201mL = 251mL = 0.251L:

2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]

As pH = -log [H+]

pH = -log 9.96x10⁻⁵M

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6 0
3 years ago
Acidic solutions contain high concentrations of
finlep [7]

Answer:

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Explanation:

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Answer:

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Explanation:

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Molar mass for the compound is:

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