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The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K areΔfG° [C(graphite)] = 0 kJ mol -1ΔfG° [C(diamond)] = 2.9 kJ mol-1The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]a)14501 barb)29001 barc)58001 bard)1405 barCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K areΔfG° [C(graphite)] = 0 kJ mol -1ΔfG° [C(diamond)] = 2.9 kJ mol-1The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]a)14501 barb)29001 barc)58001 bard)1405 barCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K areΔfG° [C(graphite)] = 0 kJ mol -1ΔfG° [C(diamond)] = 2.9 kJ mol-1The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]a)14501 barb)29001 barc)58001 bard)1405 barCorrect answer is option 'A'. Can you explain this answer?.

Solutions for The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K areΔfG° [C(graphite)] = 0 kJ mol -1ΔfG° [C(diamond)] = 2.9 kJ mol-1The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]a)14501 barb)29001 barc)58001 bard)1405 barCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K areΔfG° [C(graphite)] = 0 kJ mol -1ΔfG° [C(diamond)] = 2.9 kJ mol-1The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]a)14501 barb)29001 barc)58001 bard)1405 barCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K areΔfG° [C(graphite)] = 0 kJ mol -1ΔfG° [C(diamond)] = 2.9 kJ mol-1The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]a)14501 barb)29001 barc)58001 bard)1405 barCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K areΔfG° [C(graphite)] = 0 kJ mol -1ΔfG° [C(diamond)] = 2.9 kJ mol-1The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]a)14501 barb)29001 barc)58001 bard)1405 barCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K areΔfG° [C(graphite)] = 0 kJ mol -1ΔfG° [C(diamond)] = 2.9 kJ mol-1The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]a)14501 barb)29001 barc)58001 bard)1405 barCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298 K areΔfG° [C(graphite)] = 0 kJ mol -1ΔfG° [C(diamond)] = 2.9 kJ mol-1The standard state means that the pressure should be 1 bar , and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by 2 × 10-6 m3 mol-1. If C(graphite) is converted to C(diamond) isothermally at T = 298 K, the pressure at which C(graphite) is in equilibrium with C(diamond), is[Useful information : 1 J = 1 kg m2 s–2 ; 1 Pa = 1 kg m–1 s–2 ; 1 bar = 105 Pa]a)14501 barb)29001 barc)58001 bard)1405 barCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.