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viva [34]
3 years ago
5

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

Chemistry
1 answer:
Darya [45]3 years ago
8 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

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

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Answer:

41.3 °C

Explanation:

From the question given above, the following data were obtained:

Mass (M) of water = 27.56 g

Heat (Q) loss = 2443 J

Final temperature (T2) = 62.5 °C

Initial temperature (T1) =?

NOTE: The specific heat capacity (C) of water is 4.18 J/g°C

Thus, we can obtain the initial temperature of the water by using the following formula:

Q = MC(T2 – T1)

2443 = 27.56 × 4.18 (62.5 – T1)

2443 = 115.2008 (62.5 – T1)

Divide both side by 115.2008

2443 / 115.2008 = (62.5 – T1)

21.20645 = 62.5 – T1

Collect like terms

21.20645 – 62.5 = – T1

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Thus, the initial temperature of the water was 41.3 °C

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