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Thermodynamics question

2017 · Shift 2 · Q16
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Thermodynamics question

2017 · Shift 2 · Q16

JEE AdvancedChemistryThermodynamicsMCQ+3 / −0.75
The standard state Gibbs free energies of formation of CCC(graphite) and CCC(diamond) at T=298KT=298KT=298K are ΔfG0{\Delta _f}{G^0}Δf​G0 [CCC(graphite)] =0kJmol−1ΔfG0= 0kJmo{l^{ - 1}}{\Delta _f}{G^0}=0kJmol−1Δf​G0 [CCC(diamond)] =2.9kJmol−1= 2.9kJmo{l^{ - 1}}=2.9kJmol−1 The standard state means that the pressure should be 111 bar, and substance should be pure at a given temperature. The conversion of graphite [CCC(graphite)] to diamond [CCC(diamond)] reduces its volume by 2×10−6 m3 mol−12 \times {10^{ - 6}}\,{m^3}\,mo{l^{ - 1}}2×10−6m3mol−1 If CCC(graphite) is converted to CCC(diamond) isothermally at T=298K,T=298K,T=298K, the pressure at which CCC(graphite) is in equilibrium with CCC(diamond), is [Useful information : 1J=11J=11J=1 kg m2s−2;1 Pa=1 kg m−1s−2;kg\,{m^2}{s^{ - 2}};1\,Pa = 1\,kg\,{m^{ - 1}}{s^{ - 2}};kgm2s−2;1Pa=1kgm−1s−2; 111 bar =105Pa= {10^5}Pa=105Pa]
  1. A
    145011450114501 bar
  2. B
    58001bar58001bar58001bar
  3. C
    145014501450 bar
  4. D
    290012900129001 bar
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