If the initial concentration of NH3(g) is 4.643 mol/L, calculate the % of NH3(g) left over after the reaction reaches equilibrium according to the balanced equation. The value of Kc at 773.0 K is 16.70. The initial concentration of the reaction products is 0 mol/L.
2NH3(g) = N2(g)+3H2(g)
I know that we're supposed to do an ICE chart. I got (x^2)/ 6.745-2x . But I don't know how to solve for the x and the rest I would appreciate if you could write out the steps to this :)
You need work on your ICE charts.
...............2NH3 ==> N2 + 3H2
initial.......4.643.....0......0
change........-2x.......x......3x
equil.......4.643-2x.....x......3x
Kc = 16.70 = (N2)(H2)^3/(NH3)2
16.70 = (x)(3x)^3/(4.643-2x)^2
16.70 = 27x^4/(etc)
I didn't check my work (so I could have made a math error) but I expanded the denominator, multiplied by 16.70, and ended up with
27x^4 -66.8x^2 +310.15x -360 = 0
First I ignored the 27x^4 and solved the quadratic and obtained an answer of 2.31, substituted 2.31 back into the equation to see if I obtained zero. I didn't; the value was quite positive. Several more trials and I think the answer is >1.279 but < 1.280. You can try your hand at the trial and error method or find a computer program at school that will solve a quartic or look on the web for a calculator that will do it.
I looked on the web and found, and used, two calculators but I wasn't satisfied with either answer. One answer was 1.03 something but that didn't give me a zero when I substituted. The other one gave a value of 1.276 and that didn't give a zero either. Both showed negative numbers which means the answer is higher than 1.276.
Thank you so much :)
To solve this question using an ICE chart, follow these steps:
Step 1: Write the balanced chemical equation:
2NH3(g) = N2(g) + 3H2(g)
Step 2: Set up the ICE chart, where "I" represents the initial concentration, "C" represents the change in concentration, and "E" represents the equilibrium concentration:
| 2NH3(g) | N2(g) | 3H2(g) |
------------------------------------------
I | 4.643 | 0 | 0 |
C | -2x | +x | +3x |
E | 4.643-2x | x | 3x |
Step 3: Set up the expression for the equilibrium constant Kc using the given values:
Kc = [N2][H2]^3 / [NH3]^2
Since the initial concentrations of N2 and H2 are 0, the expression simplifies to:
Kc = [x]^3 / (4.643-2x)^2
Step 4: Substitute the known value of Kc and solve for x:
16.70 = [x]^3 / (4.643-2x)^2
Step 5: Rearrange the equation:
16.70 * (4.643-2x)^2 = [x]^3
Step 6: Expand and simplify the equation:
16.70 * (21.585-18.743x+4x^2) = x^3
Step 7: Cube both sides of the equation:
(x^3)^2 = 280.494 * (21.585-18.743x+4x^2)
Step 8: Simplify the equation:
x^6 = 280.494 * (21.585-18.743x+4x^2)
Step 9: Rearrange the equation and put it in the standard form of a cubic equation:
4x^2 - 18.743x + (21.585 - (280.494/280.494)^2) = 0
Step 10: Solve the quadratic equation for x using the quadratic formula or factoring techniques.
After calculating the value of x, you can substitute it into the ICE chart to obtain the equilibrium concentrations. Finally, calculate the percent of NH3 remaining by dividing the equilibrium concentration by the initial concentration and multiplying by 100.