Asked by ny
The synthesis reaction to form boron hydride is: 2B(s) + 3Hmc001-1.jpg(g) mc001-2.jpg Bmc001-3.jpgHmc001-4.jpg(g), mc001-5.jpgH = +36 KJ.
What is the enthalpy value for the manipulated reaction: 1/2Bmc001-6.jpgHmc001-7.jpg(g) mc001-8.jpg B(s) + 3/2Hmc001-9.jpg(g)?
–18 KJ
–36 KJ
+18 KJ
+72 KJ
All Answers 1
Answered by
GPT-5 mini
AI
Original: 2 B(s) + 3 H2(g) → B2H6(g), ΔH = +36 kJ.
Desired: 1/2 B2H6(g) → B(s) + 3/2 H2(g) is the reverse of half the original. Reverse changes the sign (→ −36 kJ) and halving multiplies by 1/2 (→ −18 kJ).
Answer: −18 kJ.
Desired: 1/2 B2H6(g) → B(s) + 3/2 H2(g) is the reverse of half the original. Reverse changes the sign (→ −36 kJ) and halving multiplies by 1/2 (→ −18 kJ).
Answer: −18 kJ.
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