Consider the following two reactions and write down the balanced chemical equations. Reaction 1: Formic acid (CH2O2, I) reacts with oxygen to produce liquid water and carbon dioxide gas. The enthalpy change for this reaction is A,H= -255 kJ mol–1 (per mole of CH2O2). Reaction 2: The formation reaction for CO2(g). The enthalpy of formation for CO2(g) is AfH= -394 k) mol-1 Given the data above, what is the enthalpy of reaction for: C(gr) + H,O(1) +0,(g) → CH, O, (1) A„H = Number kJ mol-1

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Consider the following two reactions and write down the balanced chemical equations.
Reaction 1: Formic acid (CH202, 1) reacts with oxygen to produce liquid water and carbon dioxide gas. The
enthalpy change for this reaction is A,H= – 255 kJ mol-1 (per mole of CH202).
Reaction 2: The formation reaction for CO2(g). The enthalpy of formation for CO2(g) is
AfH= - 394 kJ mol-1
Given the data above, what is the enthalpy of reaction for:
C(gr) +H,O(1) + 0,(g)
→ CH, O, (1)
Δ,Η-Number
kJ mol-1
Transcribed Image Text:Consider the following two reactions and write down the balanced chemical equations. Reaction 1: Formic acid (CH202, 1) reacts with oxygen to produce liquid water and carbon dioxide gas. The enthalpy change for this reaction is A,H= – 255 kJ mol-1 (per mole of CH202). Reaction 2: The formation reaction for CO2(g). The enthalpy of formation for CO2(g) is AfH= - 394 kJ mol-1 Given the data above, what is the enthalpy of reaction for: C(gr) +H,O(1) + 0,(g) → CH, O, (1) Δ,Η-Number kJ mol-1
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