Home › NEET UG › chemistry › c11equil_q_dg_lechatelier › Per NCERT Problem 6.10, given ΔG° = 13.8 kJ/mol …
Per NCERT Problem 6.10, given ΔG° = 13.8 kJ/mol at 298 K for glucose phosphorylation, the value of Kc is closest to which?
A5.6 × 10²
B3.8 × 10⁻³
C1.0
D3.8 × 10³
Answer & Solution
Correct answer: B. 3.8 × 10⁻³
NCERT Problem 6.10: ln Kc = −13800 / (8.314 × 298) = −5.57; Kc = e^−5.57 = 3.81 × 10⁻³.
Related questions
Per NCERT Problem 6.5, given Kc = 3.75×10⁻⁶ at 1069 K for 2 NOCl(g) ⇌ 2 NO(g) + Cl₂(g), thPer NCERT §6.8.5 + Table 6.5, for N₂ + 3H₂ ⇌ 2NH₃ (ΔH < 0), the Haber process operates at Per NCERT §6.8.5, what is the effect of adding a CATALYST on the position of equilibrium aPer NCERT §6.8.4, for an EXOTHERMIC reaction, what is the effect of raising the temperaturPer NCERT §6.8.3, adding an INERT gas (e.g. argon) at CONSTANT VOLUME to a gaseous equilibPer NCERT §6.8.2, for the equilibrium CO(g) + 3 H₂(g) ⇌ CH₄(g) + H₂O(g), an INCREASE in prPer NCERT §6.8, Le Chatelier's principle states that a stress on a system at equilibrium cPer NCERT §6.6.1, if Kc > 10³, what does this magnitude indicate about the reaction at equ