Go Figure 5.20 Part A (Figure 1) Enthalpy diagram for combustion of 1 mol of met

Question

Go Figure 5.20 Part A (Figure 1) Enthalpy diagram for combustion of 1 mol of methane. The enthalpy change of the one- step reaction equals the sum of the enthalpy changes of the reaction run in two steps: 890 kJ802 kJ +(-88 kJ). What process corresponds to the -88 kJ enthalpy change? The reaction between CO2(g) and 2H20(g) The evaporation of 2 H2O() to 2 H20(g) The condensation of 2 H20(g) to 2 H20() Submit My Answers Give Up Provide Feedback Continue Figure 1 of 1 CH4(g) +2 02(g) -802 kJ -890 kJ CO (8)+ 2 H20(g) -88 kJ CO2(g) + 2 H20(l)

Explanation / Answer

CO2 (g) + 2 H2O (g) ----------> CO2 (g) + 2 H2O (l)

So, delta H = - 88 kJ indicates for condensation of H2O (g) to H2O (l)

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