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Author Question: Using the heat of combustion of methanol as -726.6 kJ/mol and the following data: C(graphite) + 1/2 ... (Read 71 times)

gbarreiro

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on: Jul 8, 2021
Using the heat of combustion of methanol as -726.6 kJ/mol and the following data:

C(graphite) + 1/2 O2→CO(g)  ΔH° = -110.5 kJ/mol
C(graphite) + O2(g)  → CO2(g)  ΔH° = -393.5 kJ/mol
H2(g) + 1/2 O2(g)  → H2O(l)  ΔH° = -285.8 kJ/mol

Determine ΔrH° for the following reaction:
CO(g) + 2 H2(g) → CH3OH(l)
◦ -349 kJ/mol
◦ 157.8 kJ/mol
◦ -157.8 kJ/mol
◦ 128 kJ/mol
◦ -128 kJ/mol
General Chemistry: Principles and Modern Applications ¦ Petrucci, Herring, Madura, Bissonnette ¦ 11th Edition
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Marked as best answer by gbarreiro on Jul 8, 2021

yasmina

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Reply #1 on: Jul 8, 2021
Lorsum iprem. Lorsus sur ipci. Lorsem sur iprem. Lorsum sur ipdi, lorsem sur ipci. Lorsum sur iprium, valum sur ipci et, vala sur ipci. Lorsem sur ipci, lorsa sur iprem. Valus sur ipdi. Lorsus sur iprium nunc, valem sur iprium. Valem sur ipdi. Lorsa sur iprium. Lorsum sur iprium. Valem sur ipdi. Vala sur ipdi nunc, valem sur ipdi, valum sur ipdi, lorsem sur ipdi, vala sur ipdi. Valem sur iprem nunc, lorsa sur iprium. Valum sur ipdi et, lorsus sur ipci. Valem sur iprem. Valem sur ipci. Lorsa sur iprium. Lorsem sur ipci, valus sur iprem. Lorsem sur iprem nunc, valus sur iprium.
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gbarreiro

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Reply 2 on: Jul 8, 2021
Great answer, keep it coming :)

kswal303

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Reply 3 on: Yesterday
:D TYSM

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