Calculate the standard enthalpy of formation of liquid methanol, CH3OH(I), using thefollowing information:5.C (graphite) + 2H2(g) + 1/2 O2(g)CH3OH(I)Given:C (graphite) + 02 - CO2(g)H2(g) + 2 O2 H20(1)CH3OH(I) + 3/2 02(g) CO2(g) + 2H20(1)AH° = -393.5 kJ/molAH° = -285.8 kJ/molAH° = -726.4 kJ/mol6.Calculate AH for this reaction:CH4(g) + NH3(g)→HCN(g) + 3H2(g)Given:N2(g) + 3 H2(g) →2 NH3(g)C(s) + 2 H2(g) → CH4(g)H2(g) + 2 C(s) + N2(g)AH = -91.8 kJAH = -74.9 kJ→ 2 HCN(g)AH = +270.3 kJ

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Calculate the standard enthalpy of formation of liquid methanol, CH3OH(I), using the following information: 5. C (graphite) + 2H2(g) + 1/2 O2(g) → CH3OH(I) Given: C (graphite) + O2 - CO2(g) H2(g) + 2 O2 H20(1) CH:OH(I) + 3/2 02(g) - CO2(g) + 2H20(1) AH° = -393.5 kJ/mol AH° = -285.8 kJ/mol AH° = -726.4 kJ/mol

Calculate the standard enthalpy of formation of liquid methanol, CH3OH(I), using the following information: 5. C (graphite) + 2H2(g) + 1/2 O2(g) → CH3OH(I) Given: C (graphite) + O2 - CO2(g) H2(g) + 2 O2 H20(1) CH:OH(I) + 3/2 02(g) - CO2(g) + 2H20(1) AH° = -393.5 kJ/mol AH° = -285.8 kJ/mol AH° = -726.4 kJ/mol

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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