An equation that corresponds to the process of ionization enthalpy and electron affinity by using elemental phosphorus as an example is to be written. An explanation is to be given corresponds to the fact that why first ionization energy increases as one proceed from left to right. The reason is to be stated for fact that first ionization energy of Aluminum and Sulfur lower than the Magnesium and Phosphorus respectively. The reasons for the increment in the successive ionization energies are to be stated. Any large jump observed between successive ionization enthalpy of Silicon after the removal of all electrons is to stated. Concept introduction: The energy needed to eject the loosely bound electron present in an atom is called ionization energy. It is also termed as ionization enthalpy. The energy required to add an electron in an atom is called electron affinity. On-going from left to right across the period the nuclear force increases therefore, first ionization enthalpy increases. To determine: The equations for the process of ionization enthalpy and electron affinity, the reason for the increment in the first ionization enthalpy which tends to increase from left to right across a period, the reason for lower value of first ionization energy of Aluminum than Magnesium, the reason for the lower value of first ionization energy of Sulfur than phosphorous, the reason for an increment in successive ionization enthalpies of an atom, if any large jumps between successive ionization enthalpies of Silicon after the removal of all the electrons.

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MISSED THIS? Read Section 19.9 (Pages 878-881); Watch IWE 19.10 Consider the following reaction: CH3OH(g) CO(g) + 2H2(g) (Note that AG,CH3OH(g) = -162.3 kJ/mol and AG,co(g)=-137.2 kJ/mol.) Part A Calculate AG for this reaction at 25 °C under the following conditions: PCH₂OH Pco PH2 0.815 atm = 0.140 atm 0.170 atm Express your answer in kilojoules to three significant figures. Ο ΑΣΦ AG = -150 Submit Previous Answers Request Answer □? kJ × Incorrect; Try Again; 2 attempts remaining Calculate the free energy change under nonstandard conditions (AGrxn) by using the following relationship: AGrxn = AGrxn + RTInQ, AGxn+RTInQ, where AGxn is the standard free energy change, R is the ideal gas constant, T is the temperature in kelvins, a is the reaction quotient. Provide Feedback Next >

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Chapter 2, Problem 9RQ

Interpretation Introduction

Interpretation: An equation that corresponds to the process of ionization enthalpy and electron affinity by using elemental phosphorus as an example is to be written. An explanation is to be given corresponds to the fact that why first ionization energy increases as one proceed from left to right. The reason is to be stated for fact that first ionization energy of Aluminum and Sulfur lower than the Magnesium and Phosphorus respectively. The reasons for the increment in the successive ionization energies are to be stated. Any large jump observed between successive ionization enthalpy of Silicon after the removal of all electrons is to stated.

Concept introduction: The energy needed to eject the loosely bound electron present in an atom is called ionization energy. It is also termed as ionization enthalpy.

The energy required to add an electron in an atom is called electron affinity.

On-going from left to right across the period the nuclear force increases therefore, first ionization enthalpy increases.

To determine: The equations for the process of ionization enthalpy and electron affinity, the reason for the increment in the first ionization enthalpy which tends to increase from left to right across a period, the reason for lower value of first ionization energy of Aluminum than Magnesium, the reason for the lower value of first ionization energy of Sulfur than phosphorous, the reason for an increment in successive ionization enthalpies of an atom, if any large jumps between successive ionization enthalpies of Silicon after the removal of all the electrons.

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MISSED THIS? Read Section 19.9 (Pages 878-881); Watch IWE 19.10 Consider the following reaction: CH3OH(g) CO(g) + 2H2(g) (Note that AG,CH3OH(g) = -162.3 kJ/mol and AG,co(g)=-137.2 kJ/mol.) Part A Calculate AG for this reaction at 25 °C under the following conditions: PCH₂OH Pco PH2 0.815 atm = 0.140 atm 0.170 atm Express your answer in kilojoules to three significant figures. Ο ΑΣΦ AG = -150 Submit Previous Answers Request Answer □? kJ × Incorrect; Try Again; 2 attempts remaining Calculate the free energy change under nonstandard conditions (AGrxn) by using the following relationship: AGrxn = AGrxn + RTInQ, AGxn+RTInQ, where AGxn is the standard free energy change, R is the ideal gas constant, T is the temperature in kelvins, a is the reaction quotient. Provide Feedback Next >

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