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Alborosie
3 years ago
14

For the reaction

Chemistry
1 answer:
PilotLPTM [1.2K]3 years ago
8 0
We are asked to find the value of ΔG°rxn from the equilibrium concentrations of the reactants and products. We can use the following formula:

ΔG°rxn = -RTlnK

The value of R = 8.314 J/Kmol, T = 298.15 K and we are given the equilibrium constant Keq = 2.82.

The question provides equilibrium concentrations and then asks to find ΔG°rxn when more of a product is added to the reaction mixture. However, you are asked to find ΔG after the reaction has settled down and reached equilibrium once more. Therefore, we can simply use Keq = 2.82 still and solve for ΔG.

ΔG°rxn = -(8.314 J/Kmol)(298.15 K)(ln(2.82))
ΔG°rxn = -2570 J/mol
ΔG°rxn = -2.57 kJ/mol

Under equilibrium conditions at standard temperature and pressures, the value of ΔG°rxn = -2.57 kJ/mol.
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Answer:

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

Explanation:

ClO ( g ) + O_3 ( g )\rightarrow Cl ( g ) + 2 O_2 ( g ),\Delta H^o_{1,rxn} =-122.8 kJ..[1]

2 O_3 ( g )\rightarrow 3O_2 ( g ),\Delta H^o_{2,rxn} = -285.3 kJ..[2]

O_3(g) + Cl(g)\rightarrow ClO (g)+O_2(g),\Delta H^o_{3,rxn}=?..[3]

The enthalpy of reaction for the reaction of chlorine with ozone can be calculated by using Hess's law:

[2] - [1] = [3]

\Delta H^o_{3,rxn}=\Delta H^o_{2,rxn}-\Delta H^o_{1,rxn}

=-285.3 kJ-(-122.8 kJ)=162.5 kJ

The enthalpy of reaction for the reaction of chlorine with ozone is -162.5 kJ.

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3 years ago
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If we have 0.06 moles of HCI <br><br> How many grams of magnesium metal is that?
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Answer:

1.44 g

Explanation:

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

Number of mole of HCl = 0.06 mole

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From the question given above, we discovered the number of mole of HCl is equivalent to the number of mole of Mg. Thus,

Number of mole of Mg = number of mole of HCl

Number of mole of Mg = 0.06 mole

Finally, we shall determine the mass of Mg. This can be obtained as follow:

Number of mole of Mg = 0.06 mole

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Mass of Mg =?

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