Question

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When liquid benzene is oxidised at a constant pressure of 300 K, the change in enthalpy is -3728 kJ.
What is the change in internal energy at the same temperature?​

Answer 1

Answer:

The change in internal Energy (Δ U) is −3724.2587 kJ

Explanation:

Given :

Liquid $C_6H_6$ is Oxidised at constant pressure (∴Δ p = 0),

The change in energy (Δ E) = -3728 kJ (∴ Exothermic reaction)

To find :

The change in Internal energy (Δ U) at same temperature,.

Solution :

The reaction is :

⇒ $C_6H_6(l)+\frac{15}{3}O_2(g)$ ⇒ $6CO_2(g) + 3H_2O(l)$

Here,

We know that,

Δ E = - 3728kJ

So,

Here, $\Delta^{ng}$ = Number of moles of gaseous products - Number of moles of gaseous reactant

⇒ Here, the gaseous part of the reaction is :

⇒$\frac{15}{2}O_2$ ⇒ $6CO_2$

⇒ ∴$\Delta^{ng} = 6 - \frac{15}{2} = \frac{12 - 15}{2} = \frac{-3}{2}$

We know that,

The equation :

⇒$\Delta E = \Delta U + \Delta^{ng}RT$

Here,

Δ E = - 3728kJ

⇒ $\Delta^{ng} = \frac{-3}{2}$

R = 8.314 × 10⁻³ kJ/mol. K

T = 300 K

By substituting the given values,

We get,

⇒ $-3728 = \Delta U + (\frac{-3}{2}) \times (8.314 \times 10^{-3}) \times (300)$

⇒ $-3728 - \Delta U= (-450)(8.314 \times 10^{-3})$

⇒ $-\Delta U= (-450)(8.314 \times 10^{-3}) +3728$

⇒ $\Delta U = -3728 - (-450)(8.314 \times 10^{-3})$

⇒ $\Delta U = -3728 - (-3.7413)$

⇒ $\Delta U = -3728 + 3.7413$

⇒ $\Delta U = -3724.2587$

∴ The change in internal Energy (Δ U) is −3724.2587 kJ