Nitric oxide (NO) reacts readily with chlorine gas as follows: $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)$$ At \(700 \mathrm{~K},\) the equilibrium constant \(K_{p}\) for this reaction is \(2.6 \times 10^{-3}\). Predict the behavior of each of the following mixtures at this temperature and indicate whether or not the mixtures are at equilibrium. If not, state whether the mixture will need to produce more products or reactants to reach equilibrium. (a) $P_{\mathrm{NO}}=20.3 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=20.3 \mathrm{kPa}, R_{\mathrm{NOCl}}=20.3 \mathrm{kPa}$ (b) $P_{\mathrm{NO}}=25.33 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=15.2 \mathrm{kPa}, R_{\mathrm{NOCl}}=2.03 \mathrm{kPa}$ (c) $P_{\mathrm{NO}}=15.2 \mathrm{kPa}, P_{\mathrm{Cl}_{2}}=42.6 \mathrm{kPa}, P_{\mathrm{NOCl}}=5.07 \mathrm{kPa}$

Step by step solution

01

Calculate Qp for the given mixture(a)

Now, we will use the values provided to calculate the reaction quotient \(Q_p\): \(Q_p = \frac{(20.3)^2}{(20.3)^2 \times 20.3} = \frac{1}{20.3} = 0.049\)

02

Compare Qp with Kp for mixture(a)

Since \(Q_p = 0.049\) and \(K_p = 2.6 \times 10^{-3}\), we have \(Q_p > K_p\) indicating that the reaction will proceed to the left, towards reactants to reach equilibrium. (b) \(P_{NO} = 25.33 kPa\), \(P_{Cl_2} = 15.2 kPa\), \(P_{NOCl} = 2.03 kPa\)

03

Calculate Qp for the given mixture(b)

Now, we will use the values provided to calculate the reaction quotient \(Q_p\): \(Q_p = \frac{(2.03)^2}{(25.33)^2 \times 15.2} = \frac{4.1209}{9743.908} = 0.0004228\)

04

Compare Qp with Kp for mixture(b)

Since \(Q_p = 0.0004228\) and \(K_p = 2.6 \times 10^{-3}\), we have \(Q_p < K_p\), indicating that the reaction will proceed to the right, towards the products to reach equilibrium. (c) \(P_{NO} = 15.2 kPa\), \(P_{Cl_2} = 42.6 kPa\), \(P_{NOCl} = 5.07 kPa\)

05

Calculate Qp for the given mixture(c)

Now, we will use the values provided to calculate the reaction quotient \(Q_p\): \(Q_p = \frac{(5.07)^2}{(15.2)^2 \times 42.6} = \frac{25.7049}{9868.32} = 0.002605\)

06

Compare Qp with Kp for mixture(c)

Since \(Q_p = 0.002605\) and \(K_p = 2.6 \times 10^{-3}\), we have \(Q_p \approx K_p\), indicating that the reaction mixture is at equilibrium.

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