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← JEE Advanced 2026 · Paper 1
ChemistrySingle correct+3 / -1Question 34 of 48

JEE Advanced 2026 · Paper 1, Chemistry question 2

For a reversible reaction R⇌P\mathbf R\rightleftharpoons\mathbf P, at constant temperature, both the forward and the backward reactions are first order elementary reactions with rate constants kfk_f and kbk_b, respectively. At time zero, the concentration of R\mathbf R is [R]0[\mathbf R]_0 and the concentration of P\mathbf P is zero. At any given time, [R][\mathbf R] and [P][\mathbf P] are the concentrations of R\mathbf R and P\mathbf P, respectively. If kb=4kfk_b=4k_f, the correct graphical representation of the reaction is

Choose one option
Answer and worked solution

Official answer: C

At equilibrium the forward and backward rates are equal:
[P]eq[R]eq=kfkb=14.\frac{[\mathrm P]_{eq}}{[\mathrm R]_{eq}}=\frac{k_f}{k_b}=\frac14.
With [R]+[P]=[R]0[\mathrm R]+[\mathrm P]=[\mathrm R]_0: [R]eq=0.8[R]0[\mathrm R]_{eq}=0.8[\mathrm R]_0 and [P]eq=0.2[R]0[\mathrm P]_{eq}=0.2[\mathrm R]_0.

[R]/[R]0[\mathrm R]/[\mathrm R]_0 falls from 1 to 0.8 and [P]/[R]0[\mathrm P]/[\mathrm R]_0 rises from 0 to 0.2, as in graph (C).