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2 years ago

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

1 answer:

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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3 years ago

Read 2 more answers

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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Explanation:

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

As the value of $\Delta H$ is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.

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

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3 years ago