Chemistry

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Use average bond energies to calculate delta Hrxn for C(s)+2H2O(g)-->2H2(g)+CO2(g)

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(sum bond energies reactants)-(sum bond energies products)

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2H2 + O2 yield 2H2O , so if i switch it around, 2H2O yield 2H2 + O2. Would it change which one is breaking bond, which one is making bond. So technically is the left side always breaking bond, right side is forming bond
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Data: C(graphite) + O2(g) forms CO2 (ga) Delta H = -393.5kJ H2(g) + 1/2O2(g) forms H2O(l)Delta H = 285.8kJ. CH3OH(l) + 3/2O2(g)forms CO2(g) + 2H2O(l)Delta H = -726.4 Using data above, calculate the enthalpy change for the raction below …
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A quick question. When we use the bond energies to find delta H, should we use the sum of the bond energies of the reactants minus the sum of the bond energies of the products, or vice versa?
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Ok, I really need to know how to work problems like these so could someone please show me how to set it up?
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use standard enthalpies of formation to calculate delta Hrxn for the following reaction. 2H2S(g)+3O2(g)-->2H2O(l)+2SO2(g) express the answer using four significant figures Delta Hrxn = ?
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Calculate DeltaHrxn for the following reaction: C + H2O --> CO + H2 Use the following: C + O2 --> CO2 DeltaH = -393.5 kJ 2CO + O2 --> 2CO2 DeltaH = -566.0 kJ 2H2 + O2 --> 2H2O Delta H = -483.6 kJ DeltaHrxn = ?
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Calculate ΔHrxn for the following reaction: C(s)+H2O(g)→CO(g)+H2(g) Use the following reactions and given ΔH values: C(s)+O2(g)→CO2(g), ΔH= -393.5 kJ 2CO(g)+O2(g)→2CO2(g), ΔH= -566.0 kJ 2H2(g)+O2(g)→2H2O(g), …

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