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pishuonlain [190]

3 years ago

8

`as much as we would like for it to be the case, we never get all i the product we are supposed to get when we run a real reacti

on. suppose you have a theoretical yield of a 50.0 grams of product. however, your actual yield is only 32.0 g. what is your percent yield?

1 answer:

zavuch27 [327]3 years ago

3 0

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At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106

lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant ( $K_{c}$ ) for the given chemical reaction, is given by the equation:

$K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}$

<u><em>At the original equilibrium state:</em></u>

$K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}$

$K_{c} = \frac {1.6641}{0.002332} = 713.59$

<u><em>Therefore, at the new equilibrium state:</em></u>

$K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}$

$\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}$

$\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88$

$\Rightarrow [HI] = \sqrt {12.88} = 3.589 M$

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

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2 years ago