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Maksim231197 [3]
3 years ago
7

A chemist measures the energy change ΔH during the following reaction: C3H8 (g) +5O2 (g) →3CO2 (g) +4H2O (l) =ΔH−2220.kJ Use the

information to answer the following questions.
This reaction is...

endothermic.

exothermic.

Suppose
81.0g

of
C3H8

react.
Will any heat be released or absorbed?

Yes, absorbed.

Yes, released.

No.

If you said heat will be released or absorbed in the second part of this question, calculate how much heat will be released or absorbed.

kJ

Round your answer to
3

significant digits.
Chemistry
1 answer:
statuscvo [17]3 years ago
3 0

Answer:

The reaction is exothermic.

Yes, released.

The heat released is 4,08x10³ kJ.

Explanation:

For the reaction:

C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l)

The ΔH is -2220 kJ, As ΔH is <0, <em>The reaction is exothermic.</em>

As the reaction is exothermic, the heat of the reaction will be <em>released.</em>

The heat released in 81,0g is:

81,0g C₃H₈×\frac{1mol}{44,1g}×\frac{2220kJ}{1mol}= <em>4,08x10³ kJ</em>

<em>-Using molar mass of C₃H₈ to convert mass to moles and knowing that there are released 2220 kJ per mole of C₃H₈-</em>

I hope it helps!

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Height = 3.686 m

<h3>Further explanation</h3>

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Conservation of energy :

(KE+PE)₁ (downhill) = (KE+PE)₂ (up the hill)

PE₁=0⇒h=0

KE₂=0⇒v=0(stop), so equation becomes :

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Answer:

(a) The system does work on the surroundings.

(b) The surroundings do work on the system.

(c) The system does work on the surroundings.

(d) No work is done.

Explanation:

The work (W) done in a chemical reaction can be calculated using the following expression:

W = -R.T.Δn(g)

where,

R is the ideal gas constant

T is the absolute temperature

Δn(g) is the difference between the gaseous moles of products and the gaseous moles of reactants

R and T are always positive.

  • If Δn(g) > 0, W < 0, which means that the system does work on the surroundings.
  • If Δn(g) < 0, W > 0, which means that the surroundings do work on the system.
  • If Δn(g) = 0, W = 0, which means that no work is done.

<em>(a) Hg(l) ⇒ Hg(g)</em>

Δn(g) = 1 - 0 = 1. W < 0. The system does work on the surroundings.

<em>(b) 3 O₂(g) ⇒ 2 O₃(g) </em>

Δn(g) = 2 - 3 = -1. W > 0. The surroundings do work on the system.

<em>(c) CuSO₄.5H₂O(s) ⇒ CuSO₄(s) + 5H₅O(g) </em>

Δn(g) = 5 - 0 = 5. W < 0. The system does work on the surroundings.

<em>(d) H₂(g) + F₂(g) ⇒ 2 HF(g)</em>

Δn(g) = 2 - 2 = 0. W = 0. No work is done.

3 0
3 years ago
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