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pychu [463]
2 years ago
9

How much heat (kj) is required to convert 3.12 moles of liquid benzene at 75.1°c to gaseous benzene at 115.1°c?

Chemistry
1 answer:
Tems11 [23]2 years ago
3 0
The total energy includes sensible heat to raise the temperature from 75.1°C to the boiling point. It also includes the latent heat to convert the liquid to gas. Then, it also includes sensible heat from he boiling point to 115.1°C. The equation is:

Energy = nCp,liquid(T,bp - T₁) + nΔH + nCp,gas(T₂ - T,bp)
where
n is the number of moles
T,bp is the boiling point of benzene at 80.1°C
Cp,liquid = 134.8 J/mol·°C
Cp,gas = 82.44 J/mol·°C
ΔH = 87.1 J/mol

Energy = (3.12 moles)(134.8 J/mol·°C)(80.1°C - 75.1°C) + (3.12 moles)(87.1 J/mol) + (3.12 moles)(82.44 J/mol·°C)(115.1°C - 80.1°C)
Energy = 11,377.08 J
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At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate
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Since, it is shown that the reaction has been reversed. Therefore, value of K_{c} will become \frac{1}{K_{c}}.

Hence, new K_{c'} = \frac{1}{K_{c}}

                                      = \frac{1}{0.0500}

                                      = 20

Also, the number of moles of each reactant has been halved. So, K_{c''} for the reaction MgO(s) + \frac{1}{2}Cl2(g) → MgCl_{2}(s) + \frac{1}{2} O2(g) will also get halved.

Therefore,     K_{c''}  = K_{c'} = (20)^{0.5}

                               = 4.47

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As a result, K_{c} will also increase with increase in temperature.

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