The concentration of X in the gas-phase reversible reaction 2 X(g) ⇌ Y(g) is…

CHEMISTRY · Multi-part · A Level · Chemistry · Rate equations · Rate equations

Problem

The concentration of X in the gas-phase reversible reaction 2 X(g) ⇌ Y(g) is monitored at several times, and the forward rate follows the rate law

\text{rate}_{\text f}=k_{\text f}[X]^2

with

k_{\text f}=0.50\,L\,mol^{-1}\,s^{-1}

. Experimental data are summarized below.

Time / s	[X] / M	Rate_forward / $(\text{mol}\,L^{-1}\,s^{-1})$	Rate_reverse / $(\text{mol}\,L^{-1}\,s^{-1})$
0	0.400	—	0.00
20	0.355	—	0.05
40	0.320	—	0.09
60	0.300	—	—
80	0.295	—	0.14

Assume that changes in [X] between listed times occur at a constant net rate within each 20-s interval. (1) Calculate the forward rate at each time shown in the table. (2) Using the change in [X] between 40 s and 60 s, determine the reverse rate at 60 s. Clearly show how you obtain the value. (3) State the net direction of the reaction at 60 s and predict, with justification, whether the system is likely to have reached equilibrium by 80 s.

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