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# Tutor profile: Leo R.

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Leo R.
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## Questions

### Subject:Physics (Thermodynamics)

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Question:

The standard enthalpy of formation of $$H_{2}O$$(g) at 298K is −241.82 kJ mol$$^{−1}$$. Estimate its value at 100 °C given the following values of the molar heat capacities at constant pressure: $$H_{2}O$$(g): 33.58J K$$^{−1}$$mol$$^{−1}$$; $$H_{2}$$(g): 28.84J K$$^{−1}$$mol$$^{−1}$$; $$O_{2}$$(g): 29.37J K$$^{−1}$$mol$$^{−1}$$. Assume that the heat capacities are independent of temperature.The

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Leo R.

Use the integrated form of the Kirchhoff equation, $$\Delta H(T_{2})$$ = $$\Delta H(T_{1})$$ + $$\Delta Cp(T_{2} - T_{1})$$ The reaction is $$H_{2}$$(g) + $$\frac{1}{2}$$$$O_{2}$$(g) → $$H_{2}O$$(g), so $$\Delta$$Cp = Cp ($$H_{2}O$$) − {Cp ($$H_{2}$$) + $$\frac{1}{2}$$Cp ($$O_{2}$$)} = 33.58 - (28.84+$$\frac{1}{2}$$29.37) J K$$^{−1}$$mol$$^{−1}$$ = −9.94 J K$$^{−1}$$mol$$^{−1}$$ Therefore, $$\Delta H(373K)$$= −241.82kJ mol$$^{−1}$$+ (75K) × (−9.94 J K$$^{−1}$$mol$$^{−1}$$) = −242.6 kJ mol$$^{−1}$$

### Subject:Organic Chemistry

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Question:

Methlyclohexane is in rapid equilibrium between two conformers; the methyl group at the axial position and equilateral position. The difference in free energy between the two conformers is $$\Delta G = -1.4$$kcal mol$$^{-1}$$ What is the ratio of equilateral and axial methylcyclohexane in equilibrium at 298K?

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Leo R.

The question is asking the product ratio at equilibrium, K. Let's use the equation $$\Delta G =$$ -RTlnK At 298K, the equation can be rewritten as $$\Delta G =$$ -1.4logK Rearranging, K=10$$^{\frac{-1.4}{1.4}}$$ Therfore, K= 0.1

### Subject:Chemistry

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Question:

The Haber process is a reaction that produces ammonia from hydrogen and nitrogen gas. $$N_{2} + 3H_{2} <--> 2NH_{3}$$ The reaction is in equilibrium and is exothermic. What happens to the position of the equilibrium when: a) The pressure is increased? b) The temperature is decreased?

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Leo R.

To solve this problem, we must consider Le Chatelier's Principle. a) If the pressure is increased, the equilibrium will favor a state with fewer gas molecules. Since there is a total of 4 moles of reactants and 2 moles of products, the equilibrium will shift to the right. b) Since the reaction is exothermic, we can imagine "heat" as a product. So, we can write the equation as: $$N_{2} + 3H_{2} <--> 2NH_{3} +$$ Heat If the temperature is decreased, that means there is less heat. Therefore, the products are decreasing so that means the equilibrium will shift to the right.

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