Hydrogen and iodine react according to the equation H2(g) + I2(g) → 2HI(g). Suppose 1.00 mol H2 and 2.00 mol I2 are placed in a 1.00-L vessel. The number of moles of substances in the gaseous mixture when it comes to equilibrium at 458°C are equal to Blank 1. Fill in the blank, read surrounding text. mol for H2, Blank 2. Fill in the blank, read surrounding text. mol for I2, and Blank 3. Fill in the blank, read surrounding text. mol for HI. The equilibrium constant Kc at this temperature is 49.7.
The Correct Answer and Explanation is:
To solve this problem, we need to use the principles of chemical equilibrium and the equilibrium constant (Kc). The balanced reaction is:
H2(g)+I2(g)→2HI(g)\text{H}_2(g) + \text{I}_2(g) \rightarrow 2\text{HI}(g)H2(g)+I2(g)→2HI(g)
We are given the following initial conditions:
- 1.00 mol H2
- 2.00 mol I2
- 0 mol HI (since none is initially present)
The equilibrium constant, Kc, is given as 49.7 at 458°C, and the volume of the reaction vessel is 1.00 L. From these, we can set up an ICE table (Initial, Change, Equilibrium) to calculate the equilibrium concentrations of each substance.
Step 1: Set up the ICE table
| Substance | Initial (mol) | Change (mol) | Equilibrium (mol) |
|---|---|---|---|
| H2 | 1.00 | -x | 1.00 – x |
| I2 | 2.00 | -x | 2.00 – x |
| HI | 0 | +2x | 2x |
Where:
- xxx is the number of moles of H2 and I2 that react to form HI.
Step 2: Apply the equilibrium expression
The equilibrium constant KcK_cKc is given by:Kc=[HI]2[H2][I2]K_c = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]}Kc=[H2][I2][HI]2
Substitute the equilibrium concentrations into this equation:49.7=(2x)2(1.00−x)(2.00−x)49.7 = \frac{(2x)^2}{(1.00 – x)(2.00 – x)}49.7=(1.00−x)(2.00−x)(2x)2
Simplify the equation:49.7=4×2(1.00−x)(2.00−x)49.7 = \frac{4x^2}{(1.00 – x)(2.00 – x)}49.7=(1.00−x)(2.00−x)4×2
Step 3: Solve for xxx
To simplify further, multiply both sides by the denominator:49.7(1.00−x)(2.00−x)=4×249.7(1.00 – x)(2.00 – x) = 4x^249.7(1.00−x)(2.00−x)=4×2
Now expand and simplify the equation:49.7(2.00−3.00x+x2)=4×249.7(2.00 – 3.00x + x^2) = 4x^249.7(2.00−3.00x+x2)=4×299.4−149.1x+49.7×2=4×299.4 – 149.1x + 49.7x^2 = 4x^299.4−149.1x+49.7×2=4×2
Move all terms to one side:99.4−149.1x+45.7×2=099.4 – 149.1x + 45.7x^2 = 099.4−149.1x+45.7×2=0
This is a quadratic equation in the form Ax2+Bx+C=0Ax^2 + Bx + C = 0Ax2+Bx+C=0. Using the quadratic formula:x=−B±B2−4AC2Ax = \frac{-B \pm \sqrt{B^2 – 4AC}}{2A}x=2A−B±B2−4AC
where A=45.7A = 45.7A=45.7, B=−149.1B = -149.1B=−149.1, and C=99.4C = 99.4C=99.4.
Substitute the values into the quadratic formula:x=−(−149.1)±(−149.1)2−4(45.7)(99.4)2(45.7)x = \frac{-(-149.1) \pm \sqrt{(-149.1)^2 – 4(45.7)(99.4)}}{2(45.7)}x=2(45.7)−(−149.1)±(−149.1)2−4(45.7)(99.4)x=149.1±22290.81−18138.7291.4x = \frac{149.1 \pm \sqrt{22290.81 – 18138.72}}{91.4}x=91.4149.1±22290.81−18138.72x=149.1±4152.0991.4x = \frac{149.1 \pm \sqrt{4152.09}}{91.4}x=91.4149.1±4152.09x=149.1±64.4591.4x = \frac{149.1 \pm 64.45}{91.4}x=91.4149.1±64.45
Thus, we have two possible values for xxx:
- x=149.1+64.4591.4=2.33x = \frac{149.1 + 64.45}{91.4} = 2.33x=91.4149.1+64.45=2.33
- x=149.1−64.4591.4=0.93x = \frac{149.1 – 64.45}{91.4} = 0.93x=91.4149.1−64.45=0.93
We choose the solution that makes physical sense. Since we started with 1.00 mol of H2 and 2.00 mol of I2, the maximum amount that can react is 1.00 mol of H2. Thus, x=0.93x = 0.93x=0.93.
Step 4: Calculate the equilibrium concentrations
Now that we know x=0.93x = 0.93x=0.93, we can calculate the equilibrium concentrations:
- [H2] = 1.00−0.93=0.07 mol/L1.00 – 0.93 = 0.07 \, \text{mol/L}1.00−0.93=0.07mol/L
- [I2] = 2.00−0.93=1.07 mol/L2.00 – 0.93 = 1.07 \, \text{mol/L}2.00−0.93=1.07mol/L
- [HI] = 2×0.93=1.86 mol/L2 \times 0.93 = 1.86 \, \text{mol/L}2×0.93=1.86mol/L
Final Answers:
- The number of moles of H2 at equilibrium = 0.07 mol
- The number of moles of I2 at equilibrium = 1.07 mol
- The number of moles of HI at equilibrium = 1.86 mol
Thus, when the system reaches equilibrium, there will be 0.07 mol of H2, 1.07 mol of I2, and 1.86 mol of HI in the 1.00-L vessel.
