Question bank/NEET UG/Chemistry

NEET UGChemistryModerateNEET 2025

For the reaction A(g) ⇌\rightleftharpoons 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given : R = 0.0831 L atm mol⁻¹ K⁻¹]
KPK_P for the reaction at 1000 K is

Show answer and solution
Sol. KC=kfkb=12500K_C = \frac{k_f}{k_b} = \frac{1}{2500}

KP=KC(RT)ΔngK_P = K_C(RT)^{\Delta n_g} (Δng=2−1=1)(\Delta n_g = 2 - 1 = 1)

=12500×0.0831×1000= \frac{1}{2500} \times 0.0831 \times 1000

=0.033= 0.033

Source: geekyrakshit/indian-exam-questions / neet2025-code45

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