The enthalpy change (ΔH) for the reaction N2 (g) + 3H2 (g) → 2NH3

1.	The enthalpy change (ΔH) for the reaction N2 (g) + 3H2 (g) → 2NH3 (g) is -93.0 kJ at 300 K. Calculate the value of ΔE for it at 300 K. (R = 8.314 Jk-1 mol1).
Answer» ΔH = ΔE + Δn_gRT ΔH = -93 x 10³ J T = 300 K R = 8.314 Jk^-1 mol^-1 Δn_g = 2 - 4 = -2 Now ΔE = ΔH - Δn_gRT = -93 x 10³ - (-2) x 8.314 x 300 = -93000 x +4988.4 = -88011.6 J = -88.0 kJ

The enthalpy change (ΔH) for the reaction N2 (g) + 3H2 (g) → 2NH3 (g) is -93.0 kJ at 300 K. Calculate the value of ΔE for it at 300 K. (R = 8.314 Jk-1 mol1).

Answer»

ΔH = ΔE + Δn_gRT

ΔH = -93 x 10³ J

T = 300 K

R = 8.314 Jk^-1 mol^-1

Δn_g = 2 - 4 = -2

Now ΔE = ΔH - Δn_gRT

= -93 x 10³ - (-2) x 8.314 x 300

= -93000 x +4988.4

= -88011.6 J = -88.0 kJ

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The enthalpy change (ΔH) for the reaction N2 (g) + 3H2 (g) → 2NH3 (g) is -93.0 kJ at 300 K. Calculate the value of ΔE for it at 300 K. (R = 8.314 Jk-1 mol1).

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