Sum

**Answer the following question.**

Calculate ΔU at 298 K for the reaction,

C_{2}H_{4(g)} + HCl_{(g)} → C_{2}H_{5}Cl_{(g)}, ΔH = - 72.3 kJ

How much PV work is done?

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#### Solution

**Given:**

Enthalpy change = ΔH = –72.3 kJ

Temperature = T = 298 K

**To find:**

PV work done and internal energy change (ΔU)

**Formulae: **

1. W = - Δn_{g}RT

2. ΔH = ΔU + Δn_{g}RT

**Calculations:**

Δn_{g} = (moles of product gases) - (moles of reactant gases)

Δn_{g} = 1 – 2 = –1 mol

Using formula (i)

W = - Δn_{g}RT

= - (- 1 mol) × 8.314 J K^{-1} mol^{-1} × 298 K

= 2477.57 J = 2.48 kJ

Now, using formula (ii) and rearranging,

ΔU = ΔH - Δ n_{g}RT = ΔH + W = –72.3 kJ + 2.48 kJ = –69.8 kJ

∴ The PV work done is 2.48 kJ.

∴ The internal energy change (ΔU) is –69.8 kJ.

Concept: Enthalpy (H)

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