Question

the enthalpy change for the reaction n2+3h2 gives 2nh3 is -9238kj at 298K . THE INTERNAL ENERGY CHANGE AT 298K IS

Answer 1

Answer:

ΔE = - 87.425 KJ

Explanation:

Given data:

ΔH = -92.38 Kj

Temperature = 298 K

ΔE = ?

Solution:

Chemical reaction:

N₂  +3H₂  → 2 NH₃

Formula:

ΔE = ΔH + Δn RT

Δn = 2 - (3+1)

ΔE = -92380 j - ( -2 mol × 8.314 j. k⁻¹ .mol⁻¹ × 298 K)

ΔE = -92380 j - (-4955.144j)

ΔE = -92380 j + 4955.144 j

ΔE = - 87424.856 j

- 87424.856 /1000

ΔE = - 87.425 KJ

Answer 2

Given:

The reaction occurring is $N_2+3H_2\rightarrow 2NH_3$.

ΔH = -92.38kJ

Temperature = 298K

To find:

ΔU = ?

Formula to be used:

ΔU = ΔH - ΔnRT

Calculation:

ΔU = ΔH - ΔnRT

Δn = 2-4 = -2

ΔU = -92.38 - (-2RT)

ΔU = -92.38 + (2×8.314×$10^{-3}$×298)

ΔU = - 92.38 + 4.955

ΔU = -87.345kJ $mol^{-1}$

Conclusion:

The internal energy at 298 K is -87.345kJ$mol^{-1}$.

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