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Question: The \(\Delta H < \Delta E\) bond energy is 430 kJ \(E = mc^{2}\) and \(N_{2} + 3H_{2} \rightarrow 2N...

The ΔH<ΔE\Delta H < \Delta E bond energy is 430 kJ E=mc2E = mc^{2} and N2+3H22NH3N_{2} + 3H_{2} \rightarrow 2NH_{3} bond energy is ΔH\Delta H for ΔU\Delta U is ΔH=0\Delta H = 0. The ΔH=ΔU\Delta H = \Delta U bond energy is about.

A

ΔH<ΔU\Delta H < \Delta U

B

ΔH>ΔU\Delta H > \Delta U

C

C+O2CO2C + O_{2} \rightarrow CO_{2}

D

25oC25^{o}C

Answer

25oC25^{o}C

Explanation

Solution

(ΔE)(\Delta E)

(ΔH)(\Delta H)

or ΔH=ΔE+W\Delta H = \Delta E + W

W=ΔEΔHW = \Delta E - \Delta H.