enter image description here I was trying to solve question 22 above. I obtained the equation: $$50=\dfrac{4\alpha^2}{(1-\alpha)(2-\alpha)}$$ Now, the resulting quadratic is too much of a pain to solve. How would one use approximations to compute this by hand?

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    $\begingroup$ To do that, you should be able to make an educated guess. Is $\alpha$ close to 0, or close to 1, or close to 10? $\endgroup$ – Ivan Neretin Nov 1 '18 at 13:01

Let $\alpha$ = amount of $\ce{A_2}$ that reacts, so the amount of $\ce{A_2}$ that is left is 1-$\alpha$. The amount of $\ce{B_2}$ that is left is 2-$\alpha$. The amount of product that is produced is 2$\alpha$.

$$50 = \dfrac{(2\alpha)^2}{(1-\alpha)(2-\alpha)} = \dfrac{4\alpha^2}{(1-\alpha)(2-\alpha)} = \dfrac{4\alpha^2}{2-3\alpha+\alpha^2}$$

So you set the equation up correctly.

Normally I'd just bite the bullet and solve the quadratic. With K=50 that means that the products are greatly favored, and the most product that can be made is 2. So with that multiple choice the only answer that makes sense is (c) 1.866, which means that $$\alpha = 1.866/2 = 0.933$$ Checking that is quicker than solving the quadratic and checking that... $$\dfrac{(1.866)^2}{(1-0.933)(2-0.933)} = 48.7$$

Any other answer would yield an even smaller value for the equilibrium expression, so (c) must be correct.

Now solving the quadratic to determine the rounding error...

$$46\alpha^2 -150\alpha +100 = 0$$


$$\alpha = 0.93444$$

which means that amount of AB = $2(0.93444) = 1.869$

  • $\begingroup$ This is a great observation, but could you apply some sort of approximation if I remove the options? $\endgroup$ – Yusuf Hasan Nov 1 '18 at 16:22
  • $\begingroup$ I'd never trust myself to solve a quadratic by hand without checking it. // If you remove the options, then I'd assume that 50 has two significant figures (at most...). So checking 1.9 gives 37 and 1.8 gives 17, so I'd go with 1.9. // The original problem is wacky. 50 has 2 significant figures, answers (a) and (b) have 3, and answers (c) and (d) have 4. $\endgroup$ – MaxW Nov 1 '18 at 16:47
  • $\begingroup$ Why would 50 have two significant figures at most? In the form it is written, it has 1 significant figure, while in the scientific notation it can have any number of significant digits until stated otherwise (My source: chemistry.bd.psu.edu/jircitano/sigfigs.html). Also, what have you calculated as 37 and 17? $\endgroup$ – Yusuf Hasan Nov 2 '18 at 2:42
  • $\begingroup$ 50 is ambigious. Does it mean $5\cdot10^1$ ( 1 significant figure) or $5.0\cdot10^1$ ( 2 significant figures). // If AB=1.9 then K=37. If AB=1.8 then K=17. $\endgroup$ – MaxW Nov 2 '18 at 5:51
  • $\begingroup$ Can 50 also mean 5.00×10¹ ? $\endgroup$ – Yusuf Hasan Nov 2 '18 at 6:12

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