Calculate the internal energy 298K the formation of one mole of ammonia- the enthalpy change constant pressure is -42-0 kJ mol-1-Given- R-8-314J K-1- mol-1-

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N2-3H2-x2192-2NH3-x394-n-2-x2212-1-3-x2212-2-x394-H-x394-U-x394-nRT-x394-U-x394-H-x2212-x394-nRT-x394-U-x2212-42-0-xA0-kJ-mol-x2212-x2212-2-xD7-8-314-xD7-10-x2212-3-xA0-kJ-mol-K-xD7-298-xA0-K-x394-U-x2212-37-0-xA0-kJ-mol

Calculate the internal energy at $298$ K for the formation of one mole of ammonia, if the enthalpy change at constant pressure is $- 42.0$ kJ $m o l^{- 1}$ .
[Given: $R = 8.314$ J $K^{- 1}$ $m o l^{- 1}$ ]

$N_{2} + 3 H_{2} \to 2 N H_{3}$

$Δ n = 2 - (1 + 3) = - 2$

$Δ H = Δ U + Δ n R T$

$Δ U = Δ H - Δ n R T$

$Δ U = - 42.0 k J / m o l - [(- 2) \times 8.314 \times 10^{- 3} k J / m o l / K \times 298 K]$

$Δ U = - 37.0 k J / m o l$

Calculate the internal energy at 298K for the formation of one mole of ammonia, if the enthalpy change at constant pressure is −42.0 kJ mol−1.[Given: R=8.314J K−1 mol−1]

N2+3H2→2NH3Δn=2−(1+3)=−2ΔH=ΔU+ΔnRTΔU=ΔH−ΔnRTΔU=−42.0 kJ/mol−[(−2)×8.314×10−3 kJ/mol/K×298 K]ΔU=−37.0 kJ/mol

Calculate the internal energy at $298$ K for the formation of one mole of ammonia, if the enthalpy change at constant pressure is $- 42.0$ kJ $m o l^{- 1}$ .
[Given: $R = 8.314$ J $K^{- 1}$ $m o l^{- 1}$ ]

$N_{2} + 3 H_{2} \to 2 N H_{3}$

$Δ n = 2 - (1 + 3) = - 2$

$Δ H = Δ U + Δ n R T$

$Δ U = Δ H - Δ n R T$

$Δ U = - 42.0 k J / m o l - [(- 2) \times 8.314 \times 10^{- 3} k J / m o l / K \times 298 K]$

$Δ U = - 37.0 k J / m o l$