Question

question 1 (1 point)
calculate the $k_{eq}$ if all are gases.
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$$\begin{align*} \\ce{n2} + 3\\ce{h2} &\\leftrightarrows 2\\ce{nh3} + \\text{heat} \\end{align*}$$

start: \quad 0 \quad \quad \quad 0 \quad \quad \quad 1.000 \text{ atm}
equil: \quad 0.200 \text{ atm} \quad 0.600 \text{ atm} \quad 0.600 \text{ atm} \\\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad 0.400 \text{ atm}
first put in the chemical symbols, then put in the numbers. pay close attention to the coefficients
\\ k_{eq} = \frac{\\_\\_1\\_\\_^2}{\\_\\_4\\_\\_\\ \\_\\_5\\_\\_^3} = \frac{(\\_\\_2\\_\\_)^2}{(\\_\\_6\\_\\_)\\ (\\_\\_7\\_\\_)^3} = \\_\\_3\\_\\_ \\
a. $\ce{nh3}$ \quad b. $\ce{n2}$ \quad c. $\ce{h2}$ \quad d. 1 \quad e. 2 \quad f. 3 \quad g. 8.33 \quad h. 0.600 \\\\ i. 0.400 \quad j. 0.200

Explanation:

Step1: Recall \( K_{eq} \) formula for gases

For reaction \( aA + bB
ightleftharpoons cC + dD \), \( K_{eq}=\frac{[C]^c[D]^d}{[A]^a[B]^b} \) (using partial pressures for gases). Here, reaction is \( \text{N}_2 + 3\text{H}_2
ightleftharpoons 2\text{NH}_3 + \text{heat} \), so \( K_{eq}=\frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} \). Thus, 1 is \( \text{NH}_3 \) (option A), 4 is \( \text{N}_2 \) (option B), 5 is \( \text{H}_2 \) (option C).

Step2: Get equilibrium pressures

Equilibrium pressure of \( \text{NH}_3 \) is \( 0.400 \) atm (option I), so 2 is \( 0.400 \). Equilibrium pressure of \( \text{N}_2 \) is \( 0.200 \) atm (option J), so 6 is \( 0.200 \). Equilibrium pressure of \( \text{H}_2 \) is \( 0.600 \) atm (option H), so 7 is \( 0.600 \).

Step3: Calculate \( K_{eq} \)

Substitute into formula: \( K_{eq}=\frac{(0.400)^2}{(0.200)(0.600)^3} \). Calculate numerator: \( (0.400)^2 = 0.16 \). Denominator: \( (0.200)(0.216)=0.0432 \). Then \( K_{eq}=\frac{0.16}{0.0432}\approx 3.70 \)? Wait, no, wait: Wait, equilibrium pressures: Wait, the reaction starts with \( \text{NH}_3 = 1.000 \) atm, equil is \( 0.600 \)? Wait no, the equil for \( \text{NH}_3 \): start is 1.000, equil is 0.600? Wait no, the table: start: \( \text{N}_2=0 \), \( \text{H}_2=0 \), \( \text{NH}_3=1.000 \) atm. Equil: \( \text{N}_2=0.200 \), \( \text{H}_2=0.600 \), \( \text{NH}_3=0.600 \)? Wait no, the second line for equil: \( \text{NH}_3 \) has two values? Wait, maybe typo, but the first equil line for \( \text{NH}_3 \) is 0.600? Wait no, the user's image: equil: \( \text{N}_2=0.200 \), \( \text{H}_2=0.600 \), \( \text{NH}_3=0.600 \) (first) and 0.400 (second)? Wait, no, probably the correct equil for \( \text{NH}_3 \) is 0.400 (since start is 1.000, change is -0.400, so \( \text{N}_2 \) change is +0.200, \( \text{H}_2 \) change is +0.600, which matches coefficients 1:3:2). So \( \text{NH}_3 \) equil is 0.400. So recalculate: \( K_{eq}=\frac{(0.400)^2}{(0.200)(0.600)^3} \). \( 0.4^2 = 0.16 \). \( 0.6^3 = 0.216 \). \( 0.2 * 0.216 = 0.0432 \). \( 0.16 / 0.0432 ≈ 3.70 \). But the options: Wait, maybe I misread the equil pressures. Wait, the equil for \( \text{NH}_3 \): maybe the first equil is 0.600, but that would mean the reaction proceeds to form \( \text{N}_2 \) and \( \text{H}_2 \). So \( \text{NH}_3 \) decomposes: \( 2\text{NH}_3
ightleftharpoons \text{N}_2 + 3\text{H}_2 \). So the reaction is reverse: \( \text{N}_2 + 3\text{H}_2
ightleftharpoons 2\text{NH}_3 \) is reverse, so \( K_{eq} \) for reverse is \( 1/K \) of forward. Wait, maybe the equilibrium pressures: \( \text{N}_2 = 0.200 \), \( \text{H}_2 = 0.600 \), \( \text{NH}_3 = 0.600 \). Wait, let's recalculate with \( \text{NH}_3 = 0.600 \). Then \( K_{eq}=\frac{(0.600)^2}{(0.200)(0.600)^3}=\frac{1}{(0.200)(0.600)}=\frac{1}{0.12}≈8.33 \). Ah! That must be it. So equil \( \text{NH}_3 \) is 0.600 (option H), not 0.400. So my mistake earlier: the equil for \( \text{NH}_3 \) is 0.600 (the first value in equil row). So \( \text{NH}_3 \) equil pressure is 0.600 (option H), so 2 is 0.600. Then numerator: \( (0.600)^2 = 0.36 \). Denominator: \( (0.200)(0.600)^3 = (0.200)(0.216) = 0.0432 \). Then \( K_{eq}=\frac{0.36}{0.0432}=8.33 \) (option G). So 3 is 8.33 (option G).

Answer:

1: A. \( \text{NH}_3 \)
2: H. \( 0.600 \)
3: G. \( 8.33 \)
4: B. \( \text{N}_2 \)
5: C. \( \text{H}_2 \)
6: J. \( 0.200 \)
7: H. \( 0.600 \)

(Note: The key was realizing the equilibrium pressure of \( \text{NH}_3 \) is 0.600 atm, not 0.400, as the first value in the equil row for \( \text{NH}_3 \) is 0.600, and the 0.400 might be a typo or misread. With \( \text{NH}_3 = 0.600 \), the calculation gives \( K_{eq} = 8.33 \), which matches option G.)