Calculate the pH at which the following conversion(reaction) will be at equilibrium in basic medium ?When the equilibrium concentrations at 300 K are [I-] = 0.10 M and [IO3-] = 0.10 M(state) = 0, andDelta;Granddeg;(reaction) = andsum;vp andDelta;Gfanddeg;(products) - andsum;vr andDelta;Ganddeg;(reactants), where vp and vr are the stochiometric coefficients in the balanced chemical equation.}{Given that andDelta;Gfanddeg;(I-,aq) = - 50 kJ/mole , andDelta; Gf0(IO3-,a q) = - 123.5 kJ/mole , andDelta;Gf0(H2O, ) = - 233 kJ/mole ,andDelta;Gf0(HO-,aq) = - 150 kJ/mole, Ideal gas constant = R = 25/3Jmole-1K-1, log e = 2.3, andDelta;Gfanddeg;{element, standard}Correct answer is '8'. Can you explain this answer?

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Calculate the pH at which the following conversion(reaction) will be at equilibrium in basic medium ?

When the equilibrium concentrations at 300 K are [I^-] = 0.10 M and [IO₃^-] = 0.10 M

(state) = 0, ΔG_r°(reaction) = ∑v_p ΔG_f°(products) - ∑v_r ΔG°(reactants), where v_p and v_r are the stochiometric coefficients in the balanced chemical equation.}
{Given that ΔG_f°(I^-,aq) = - 50 kJ/mole , Δ G_f⁰(IO₃^-,a q) = - 123.5 kJ/mole , ΔG_f⁰(H₂O, ℓ) = - 233 kJ/mole ,
ΔG_f⁰(HO^- ,aq) = - 150 kJ/mole, Ideal gas constant = R = 25/3 Jmole^-1K^-1, log e = 2.3, ΔG_f°{element, standard}

Correct answer is '8'. Can you explain this answer?

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Calculate the pH at which the following conversion(reaction) will be a...

Balanced equation will be

so [OH^-] = 10-6. and therefore [H⁺] = 10^-8 so, pH = 8 Ans.

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Calculate the pH at which the following conversion(reaction) will be at equilibrium in basic medium ?When the equilibrium concentrations at 300 K are [I-] = 0.10 M and [IO3-] = 0.10 M(state) = 0, ΔGr°(reaction) = ∑vp ΔGf°(products) - ∑vr ΔG°(reactants), where vp and vr are the stochiometric coefficients in the balanced chemical equation.}{Given that ΔGf°(I-,aq) = - 50 kJ/mole , Δ Gf0(IO3-,a q) = - 123.5 kJ/mole , ΔGf0(H2O, ) = - 233 kJ/mole ,ΔGf0(HO-,aq) = - 150 kJ/mole, Ideal gas constant = R = 25/3Jmole-1K-1, log e = 2.3, ΔGf°{element, standard}Correct answer is '8'. Can you explain this answer?

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Calculate the pH at which the following conversion(reaction) will be at equilibrium in basic medium ?When the equilibrium concentrations at 300 K are [I-] = 0.10 M and [IO3-] = 0.10 M(state) = 0, ΔGr°(reaction) = ∑vp ΔGf°(products) - ∑vr ΔG°(reactants), where vp and vr are the stochiometric coefficients in the balanced chemical equation.}{Given that ΔGf°(I-,aq) = - 50 kJ/mole , Δ Gf0(IO3-,a q) = - 123.5 kJ/mole , ΔGf0(H2O, ) = - 233 kJ/mole ,ΔGf0(HO-,aq) = - 150 kJ/mole, Ideal gas constant = R = 25/3Jmole-1K-1, log e = 2.3, ΔGf°{element, standard}Correct answer is '8'. 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 Calculate the pH at which the following conversion(reaction) will be at equilibrium in basic medium ?When the equilibrium concentrations at 300 K are [I-] = 0.10 M and [IO3-] = 0.10 M(state) = 0, ΔGr°(reaction) = ∑vp ΔGf°(products) - ∑vr ΔG°(reactants), where vp and vr are the stochiometric coefficients in the balanced chemical equation.}{Given that ΔGf°(I-,aq) = - 50 kJ/mole , Δ Gf0(IO3-,a q) = - 123.5 kJ/mole , ΔGf0(H2O, ) = - 233 kJ/mole ,ΔGf0(HO-,aq) = - 150 kJ/mole, Ideal gas constant = R = 25/3Jmole-1K-1, log e = 2.3, ΔGf°{element, standard}Correct answer is '8'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the pH at which the following conversion(reaction) will be at equilibrium in basic medium ?When the equilibrium concentrations at 300 K are [I-] = 0.10 M and [IO3-] = 0.10 M(state) = 0, ΔGr°(reaction) = ∑vp ΔGf°(products) - ∑vr ΔG°(reactants), where vp and vr are the stochiometric coefficients in the balanced chemical equation.}{Given that ΔGf°(I-,aq) = - 50 kJ/mole , Δ Gf0(IO3-,a q) = - 123.5 kJ/mole , ΔGf0(H2O, ) = - 233 kJ/mole ,ΔGf0(HO-,aq) = - 150 kJ/mole, Ideal gas constant = R = 25/3Jmole-1K-1, log e = 2.3, ΔGf°{element, standard}Correct answer is '8'. Can you explain this answer?.

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