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Nataly [62]
3 years ago
14

What is the standard enthalpy change for the following reaction?

Chemistry
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

- 2977 kJ

Explanation:

Let's consider the following reaction.

3 CH₄(g) + 4 O₃(g) → 3 CO₂(g) + 6 H₂O(g)

We can find the standard enthalpy change for the reaction (ΔH°r) using the following expression.

ΔH°r = [3 mol × ΔH°f(CO₂(g)) + 6 mol × ΔH°f(H₂O(g))] - [3 mol × ΔH°f(CH₄(g)) + 4 mol × ΔH°f(O₃(g))]

ΔH°r = [3 mol × (-393.5 kJ/mol) + 6 mol × (-241.8 kJ/mol)] - [3 mol × (-74.87 kJ/mol) + 4 mol × (142.7 kJ/mol)]

ΔH°r = - 2977 kJ

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Answer: a) pH = 13.00 : basic

b) [H_3O^+]=1.0\times 10^{-12}: basic

c) pOH = 5.00 : basic

d) [OH^-]=1.0\times 10^{-9}: acidic

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH=-\log [H_3O^+]

Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.

a) pH = 13.00

As pH is more than 7, the solution is basic.

b)  [H_3O^+]=1.0\times 10^{-12}

Putting in the values:

pH=-\log[1.0\times 10^{-12}]

pH=12

As pH is more than 7, the solution is basic.

c) pOH = 5.00

pH+pOH=14.0

pH=14.0-5.00=9.00

As pH is more than 7, the solution is basic.

d) [OH^-]=1.0\times 10^{-9}

Putting in the values:

pOH=-\log[1.0\times 10^{-9}]

pOH=9.00

pH+pOH=14.0

pH=14.0-9.00=5.00

As pH is less than 7, the solution is acidic.

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By way of example, chemists and physicists habitually use

P

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∘

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P

SVP

=

1

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a

t

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