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3 years ago

Calculate delta g reaction at 298 k under the conditions 2hg+o2

1 answer:

4 0

The balanced chemical reaction is 2Hg + O2 -> 2HgO. To calculate the delta G, we use the formula: ΔG = ΔH – TΔS. The ΔH is computed by subtracting the standard enthalpy of formation of the reactants from the products, multiplied correspondingly with their stoichiometric coefficients. The same is to obtain ΔS.

The necessary information all at 298 K are:

ΔH_Hg = 0 kJ/mol; since it is at its standard state

ΔS_Hg = 75.90 kJ/mol

ΔH_O2 = 0 kJ/mol; since it is at its standard state

ΔS_O2 = 0.161 kJ/mol K

ΔH_HgO = -90 kJ/mol

ΔS_HgO = 0.070 kJ/mol K

Thus, the delta G of reaction is -132.022 kJ. The negative value denotes a spontaneous reaction.

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3 years ago

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3 years ago

Consider the reaction: N2(g) + O2(g) ⇄ 2NO(g) Kc = 0.10 at 2000oC Starting with initial concentrations of 0.040 mol/L of N2 and

IrinaVladis [17]

Answer:

0.011 mol/L

Explanation:

This can be solved with something called an ICE table.

I = initial

C = change

E = equilibrium

Initially, there is 0.04 M of N₂, 0.04 M of O₂, and 0 M of NO.

x amount of N₂ reacts. Since the stoichiometry is 1:1, x amount of O₂ also reacts. This produces 2x of NO.

After the reaction, there is 0.04-x of N₂, 0.04-x of O₂, and 2x of NO.

Here it is in table form:

$\left[\begin{array}{cccc}&N2&O2&NO\\I&0.04&0.04&0\\C&-x&-x&+2x\\E&0.04-x&0.04-x&2x\end{array}\right]$

Now we can use the equilibrium constant:

Kc = [NO]² / ( [N₂] [O₂] )

Substituting:

0.10 = (2x)² / ( (0.04 - x) (0.04 - x) )

Solving:

0.10 = (2x)² / (0.04 - x)²

√0.10 = 2x / (0.04 - x)

(√0.10) (0.04 - x) = 2x

(√0.10)(0.04) - (√0.10)x = 2x

(√0.10)(0.04) = 2x + (√0.10)x

(√0.10)(0.04) = (2 + √0.10)x

x = (√0.10)(0.04) / (2 + √0.10)

x = 0.0055

At equilibrium, the concentration of NO is 2x. So the answer is:

[NO] = 2x

[NO] = 0.011

The equilibrium concentration of NO is 0.011 mol/L.

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