A student investigates the reversible reaction at by continuously measuring the rates…

CHEMISTRY · Multi-part · CBSE · Chemistry · Equilibrium · Equilibrium

Problem

A student investigates the reversible reaction 2 X(g)⇌Y(g)2\,X(g) \rightleftharpoons Y(g) at 298 K298\,\text{K} by continuously measuring the rates of the forward and reverse reactions. The student produces a line graph (not shown) of rate (mol L−1 s−1\text{mol}\,\text{L}^{-1}\,\text{s}^{-1}) versus time (s\text{s}) with the following key features: • At t=0 st = 0\,\text{s}, the forward rate is 0.080 and decreases smoothly, leveling off at 0.020 by about 400 s400\,\text{s}. • The reverse rate is 0.000 at t=0 st = 0\,\text{s} and increases smoothly, reaching 0.020 by about 400 s400\,\text{s}. • For t>400 st > 400\,\text{s}, the two curves coincide and remain flat at 0.020. Answer the following questions about the reaction mixture. (a) Identify the approximate time at which the system first reaches equilibrium and cite the graphical evidence that supports your answer. (b) For the interval 0 s<t<400 s0\,\text{s} < t < 400\,\text{s}, describe the direction of the net chemical change in the system and explain your reasoning in terms of the relative magnitudes of the two rate curves. (c) For t>400 st > 400\,\text{s}, explain why no further net change in composition is expected even though both reactions continue to occur. (d) Suppose the experiment is repeated at a higher temperature, and both rate curves intersect at 250 s250\,\text{s} instead of 400 s400\,\text{s}. Predict how the initial forward rate at the higher temperature would compare with 0.080 mol L−1 s−10.080\,\text{mol}\,\text{L}^{-1}\,\text{s}^{-1}, and justify your prediction.

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