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3 years ago

Enthalpy of atomisation

Chemistry

1 answer:

Softa [21]3 years ago

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Answer:

The enthalpy of atomization is the enthalpy change that accompanies the total separation of all atoms in a chemical substance. This is often represented by the symbol ΔₐₜH or ΔHₐₜ. All bonds in the compound are broken in atomization and none are formed, so enthalpies of atomization are always positive.

Explanation:

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Given the following balanced equation, determine the rate of reaction with respect to [SO3]. If the rate of O2 loss is 3.56 x 10

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Answer:

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Explanation:

Step 1: Write the balanced reaction

2 SO₂ + O₂ ⇒ 2 SO₃

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What are the concentrations of A , A, B , B, and C C at equilibrium if, at the beginning of the reaction, their concentrations a

Elenna [48]

The question is incomplete, here is the complete question:

A reaction

$A+B\rightleftharpoons C$

has a standard free-energy change of -4.88 kJ/mol at 25°C

What are the concentrations of A, B, and C at equilibrium if at the beginning of the reaction their concentrations are 0.30 M, 0.40 M and 0 M respectively?

<u>Answer:</u> The equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

<u>Explanation:</u>

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K_c$

where,

$\Delta G^o$ = Standard Gibbs free energy = -4.88 kJ/mol = -4880 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

$K_c$ = equilibrium constant of the reaction

Putting values in above equation, we get:

$-4880J/mol=-(8.3145J/Kmol)\times 298K\times \ln K_c\\\\K_c=7.17$

We are given:

Initial concentration of A = 0.30 M

Initial concentration of B = 0.40 M

Initial concentration of C = 0 M

The chemical reaction follows:

$A+B\rightleftharpoons C$

<u>Initial:</u> 0.30 0.40 0

<u>At eqllm:</u> 0.30-x 0.40-x x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[C]}{[A][B]}$

We are given:

$K_c=7.17$

Putting values in above equation, we get:

$7.17=\frac{x}{(0.30-x)\times (0.40-x)}\\\\x=0.183,0.657$

Neglecting the value of x = 0.657, because change cannot be greater than the initial concentration

So, equilibrium concentration of A = $(0.30-x)=(0.30-0.183)=0.117M$

Equilibrium concentration of B = $(0.40-x)=(0.40-0.183)=0.217M$

Equilibrium concentration of C = $x=0.183M$

Hence, the equilibrium concentrations of A, B and C are 0.117 M, 0.217 M, 0.183 M respectively.

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