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Question: Calculate ΔΗ for following reaction, at 25°C. $NH_2CN_{(g)} + \frac{3}{2}O_{2(g)} \longrightarrow N...

Calculate ΔΗ for following reaction, at 25°C.

NH2CN(g)+32O2(g)⟶N2(g)+CO2(g)NH_2CN_{(g)} + \frac{3}{2}O_{2(g)} \longrightarrow N_{2(g)} + CO_{2(g)}

(ΔU=−740.5\Delta U = -740.5 kJ, R = 8.314 J K−1K^{-1} mol)

Answer

ΔH ≈ -741.7 kJ

Explanation

Solution

Given the reaction:

NH2CN(g)+32O2(g)⟶N2(g)+CO2(g)NH_2CN_{(g)} + \frac{3}{2}O_{2(g)} \longrightarrow N_{2(g)} + CO_{2(g)}

  1. Count moles of gases:

    • Reactants:
      • NH2CNNH_2CN: 1 mole
      • O2O_2: 1.5 moles
      • Total = 2.5 moles
    • Products:
      • N2N_2: 1 mole
      • CO2CO_2: 1 mole
      • Total = 2 moles

    Change in moles:

    Δn=2−2.5=−0.5 moles\Delta n = 2 - 2.5 = -0.5 \text{ moles}

  2. Use the relation between ΔH and ΔU:

    ΔH=ΔU+Δ(PV)\Delta H = \Delta U + \Delta(PV)

    For ideal gases,

    Δ(PV)=Δn RT\Delta(PV) = \Delta n \, RT

    Given:

    • ΔU=−740.5 kJ\Delta U = -740.5\, \text{kJ}
    • T=298 KT = 298\, \text{K} (25°C)
    • R=8.314 J K−1mol−1=0.008314 kJ K−1mol−1R = 8.314\, \text{J\,K}^{-1}\text{mol}^{-1} = 0.008314\, \text{kJ\,K}^{-1}\text{mol}^{-1}

  3. Calculation:

    Δ(PV)=−0.5×0.008314 kJK mol×298 K≈−1.2385 kJ\Delta(PV) = -0.5 \times 0.008314\, \frac{\text{kJ}}{\text{K mol}} \times 298\, \text{K} \approx -1.2385\, \text{kJ}

    Thus,

    ΔH=−740.5 kJ−1.2385 kJ≈−741.74 kJ\Delta H = -740.5\, \text{kJ} - 1.2385\, \text{kJ} \approx -741.74\, \text{kJ}

Answer:

ΔH≈−741.7 kJ\Delta H \approx -741.7\, \text{kJ}