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Exercise · Q21

Q.For the reaction N2(g)+3H2(g)→2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g) at 298 K298\ \text{K}, ΔU∘=−91.8 kJ\Delta U^\circ = -91.8\ \text{kJ}. Calculate ΔH∘\Delta H^\circ for the reaction. (R=8.314 J K−1mol−1R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}.)

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For N2(g)+3H2(g)→2NH3(g)\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g), the change in moles of gas is Δng=(moles gaseous products)−(moles gaseous reactants)=2−(1+3)=2−4=−2\Delta n_g = (\text{moles gaseous products}) - (\text{moles gaseous reactants}) = 2 - (1+3) = 2 - 4 = -2. Using ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RT: ΔngRT=(−2)(8.314)(298)=−4955.1 J=−4.955 kJ\Delta n_g RT = (-2)(8.314)(298) = -4955.1\ \text{J} = -4.955\ \text{kJ}. So ΔH∘=ΔU∘+ΔngRT=−91.8+(−4.955)=−96.76 kJ\Delta H^\circ = \Delta U^\circ + \Delta n_g RT = -91.8 + (-4.955) = -96.76\ \text{kJ}. Since the reaction consumes more moles of gas than it produces, the system …

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