A2( g)+3 B2( g)→2AB3( g) Δn(g)=n(P)−n(R) =2−3−1=−2 ΔH=ΔU+ΔngRT 15×1000=ΔU−2×8.314×300 ΔU=15000+600×8.314 =15000+6×831.4 =15000+4988.4 ΔU=19988.4 J
For the following reaction at 300 K A2( g)+3 B2( g)→2AB3( g) the enthalpy change is +15 kJ then the internal energy change is:
Held on 30 Apr 2024 · Verified 9 Jul 2026.
19988.4 J
200 J
1999 J
1.9988 kJ
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