Predicting the Direction of a Reaction by Comparing Q and K Consider the reaction and its equilibrium constant. [latex]I_2(g) + Cl_2(g)\rightleftharpoons 2 \ ICl(g) [/latex] [latex]K_p = 81.9[/latex] A reaction mixture contains [latex]P_{I_2} = 0.114 \ atm, P_{Cl_2} = 0.102 \ atm,[/latex] and [latex]P_{ICl} = 0.355[/latex] atm. Is the reaction mixture at equilibrium? If not, in which direction will the reaction proceed?

To determine the progress of the reaction relative to the equilibrium state, first calculate [latex]\mathcal{Q} [/latex]. [latex]\mathcal{Q_p} =\left(\frac{P^2_{ICI}}{P_{I_2}P_{CI_2}}\right)[/latex] [latex]=\frac{(0.355)^2}{(0.114)(0.102)} [/latex] = 10.8 Compare [latex]\mathcal{Q} [/latex] to K. [latex]\mathcal{Q_p} = 10.8; K_p = 81.9 [/latex] Since [latex]\mathcal{Q_p} < K_p,[/latex] the reaction is not at equilibrium and will proceed to the right.

Question 15.7: Predicting the Direction of a Reaction by Comparing Q and K ...

Chemistry: Structure and Properties; Mastering Chemistry

Predicting the Direction of a Reaction by Comparing Q and K

Consider the reaction and its equilibrium constant.

$I_2(g) + Cl_2(g)\rightleftharpoons 2 \ ICl(g)$ $K_p = 81.9$

A reaction mixture contains $P_{I_2} = 0.114 \ atm, P_{Cl_2} = 0.102 \ atm,$ and $P_{ICl} = 0.355$ atm. Is the reaction mixture at equilibrium? If not, in which direction will the reaction proceed?

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