Question

Calculate the standard enthalpy change for the reaction 2C 8H 18(l) + 21O 2(g) ? 8CO(g) + 8CO 2(g) + 18H 2O(l). Given: 2C 8H 18(l) + 25O 2(g) ? 16CO 2(g) + 18H 2O(l) ?H° = -11,020 kJ/mol 2CO(g) + O 2(g) ? 2CO 2(g) ?H° = -566.0 kJ/mol -1.0454 × 10^4 kJ/mol -8,756 kJ/mol -6,492 kJ/mol 1.1586 × 10^4 kJ/mol 1.0454 × 10^4 kJ/mol

Name: Calculate the standard enthalpy change for the reaction 2C 8H 18(l) + 21O 2(g) ? 8CO(g) + 8CO 2(g) + 18H 2O(l). Given: 2C 8H 18(l) + 25O 2(g) ? 16CO 2(g) + 18H 2O(l) ?H° = -11,020 kJ/mol 2CO(g) + O 2(g) ? 2CO 2(g) ?H° = -566.0 kJ/mol -1.0454 × 10^4 kJ/mol -8,756 kJ/mol -6,492 kJ/mol 1.1586 × 10^4 kJ/mol 1.0454 × 10^4 kJ/mol
Uploaded: 2022-08-10T07:45:43-08:00
Duration: 4 min 13 s
Channel: Dominique Jan Tan
Description: Calculate the standard enthalpy change for the reaction 2C 8H 18(l) + 21O 2(g) ? 8CO(g) + 8CO 2(g) + 18H 2O(l). Given: 2C 8H 18(l) + 25O 2(g) ? 16CO 2(g) + 18H 2O(l) ?H° = -11,020 kJ/mol 2CO(g) + O 2(g) ? 2CO 2(g) ?H° = -566.0 kJ/mol -1.0454 × 10^4 kJ/mol -8,756 kJ/mol -6,492 kJ/mol 1.1586 × 10^4 kJ/mol 1.0454 × 10^4 kJ/mol

          Calculate the standard enthalpy change for the reaction 2C 8H 18(l) + 21O 2(g) ? 8CO(g) + 8CO 2(g) + 18H 2O(l). Given: 2C 8H 18(l) + 25O 2(g) ? 16CO 2(g) + 18H 2O(l) ?H° = -11,020 kJ/mol 2CO(g) + O 2(g) ? 2CO 2(g) ?H° = -566.0 kJ/mol -1.0454 × 10^4 kJ/mol -8,756 kJ/mol -6,492 kJ/mol 1.1586 × 10^4 kJ/mol 1.0454 × 10^4 kJ/mol

Calculate the standard enthalpy change for the reaction 2C 8H 18(l) + 21O 2(g) ? 8CO(g) + 8CO 2(g) + 18H 2O(l). Given: 2C 8H 18(l) + 25O 2(g) ? 16CO 2(g) + 18H 2O(l) ?H° = -11,020 kJ/mol 2CO(g) + O 2(g) ? 2CO 2(g) ?H° = -566.0 kJ/mol -1.0454 × 10^4 kJ/mol -8,756 kJ/mol -6,492 kJ/mol 1.1586 × 10^4 kJ/mol 1.0454 × 10^4 kJ/mol

Added by Stacy G.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Dominique Jan Tan

08/10/2022

Step 1

The standard enthalpy change for this reaction is -2*(-5566.0 kJ/mol) = 11132.0 kJ/mol. Show more…

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Calculate the standard enthalpy change for the reaction 2C8H18(l) + 21O2(g) → 8CO(g) + 8CO2(g) + 18H2O(l). Given: 2C8H18(l) + 25O2(g) → 16CO2(g) + 18H2O(l) ΔH° = -11,020 kJ/mol. 2CO(g) + O2(g) → 2CO2(g) ΔH° = -566.0 kJ/mol. -1.0454 x 10^4 kJ/mol. -8,756 kJ/mol. -6,492 kJ/mol. 1.1586 x 10^4 kJ/mol. 1.0454 x 10^4 kJ/mol.