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3 years ago

A reaction begins with 1.00 mol of HCl in a 4.00 liter container at a certain temperature.

Chemistry

1 answer:

lubasha [3.4K]3 years ago

3 0

Answer:

$[HCl]_{eq}=0.05M$

Explanation:

Hello,

In this case, for the given chemical reaction, the law of mass action at equilibrium results:

$Kc=\frac{[H_2]_{eq}[Cl_2]_{eq}}{[HCl]^2_{eq}}$

Next, in terms of the change $x$ due to reaction extent, it is rewritten, considering an initial concentration of HCl of 0.25M (1mol/4L), as:

$4.00=\frac{(x)(x)}{(0.25-2x)^2}$

Thus, solving for $x$ via quadratic equation or solver, the following results are obtained:

$x_1=0.1M\\x_2=0.17M$

Clearly, the solution is $x=0.1M$ as the other result will provide a negative concentration for the hydrochloric acid at equilibrium, thereby, its equilibrium concentration turns out:

$[HCl]_{eq}=0.25M-2*0.1M$

$[HCl]_{eq}=0.05M$

Best regards.

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An ethylene glycol solution contains 21.4 g of ethylene glycol (C2H6O2) in 97.6 mL of water. (Assume a density of 1.00 g/mL for

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Answer: The freezing point and boiling point of the solution are $-6.6^0C$ and $101.8^0C$ respectively.

Explanation:

Depression in freezing point:

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where,

$T_f$ = freezing point of solution = ?

$T^o_f$ = freezing point of water = $0^0C$

$k_f$ = freezing point constant of water = $1.86^0C/m$

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$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(0-T_f)^0C=1\times (1.86^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_f=-6.6^0C$

Therefore,the freezing point of the solution is $-6.6^0C$

Elevation in boiling point :

$T_b-T^b^0=i\times k_b\times \frac{w_2\times 1000}{M_2\times w_1}$

where,

$T_b$ = boiling point of solution = ?

$T^o_b$ = boiling point of water = $100^0C$

$k_b$ = boiling point constant of water = $0.52^0C/m$

i = vant hoff factor = 1 ( for non electrolytes)

m = molality

$w_2$ = mass of solute (ethylene glycol) = 21.4 g

$w_1$ = mass of solvent (water) = $density\times volume=1.00g/ml\times 97.6ml=97.6g$

$M_2$ = molar mass of solute (ethylene glycol) = 62g/mol

Now put all the given values in the above formula, we get:

$(T_b-100)^0C=1\times (0.52^0C/m)\times \frac{(21.4g)\times 1000}{97.6g\times (62g/mol)}$

$T_b=101.8^0C$

Thus the boiling point of the solution is $101.8^0C$

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Mole of KC₂H₃O₂ = 0.9625 mole

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