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3 years ago

What is the numerical value of Kc for the following reaction if the equilibrium mixture contains 0.51 M

Chemistry

2 answers:

MatroZZZ [7]3 years ago

7 0

Answer:- Kc = 11294

Solution:- The given balanced equation is:

$C_3H_6O(g)+4O_2(g)\rightleftharpoons 3CO_2(g)+3H_2O(g)$

Let's write the equilibrium expression for this:

$Kc=\frac{[CO_2]^3[H_2O]^3}{[C_3H_6O][O_2]^4}$

Let's plug in the values of given equilibrium concentrations and do the calculations:

$Kc=\frac{[1.8]^3[2.0]^3}{[0.51][0.30]^4}$

Kc = 11294

So, the numerical value of Kc is 11294.

Goryan [66]3 years ago

3 0

The numerical value of Kc is 1.129 x10^4

Explanation

C3H6O +4O2→ 3 CO2 + 3H2O

KC is the ratio of concentration of the product over the reactant.

Each concentration of product and reactant are raised to the power of its coefficient.

Therefore the KC expression of equation above is

Kc=[ (Co2)^3 (H2O)^3] / [(C3H6O) (O2)^4)]

Kc =[(1.8^3) x (2.0^3)] / [(0.51) x (0.30^4)] =1.129 x10^4

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jek_recluse [69]

The question is incomplete, the complete question is:

A certain substance X has a normal freezing point of $-6.4^oC$ and a molal freezing point depression constant $K_f=3.96^oC.kg/mol$ . A solution is prepared by dissolving some glycine in 950. g of X. This solution freezes at $-13.6^oC$ . Calculate the mass of urea that was dissolved. Round your answer to 2 significant digits.

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Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

$\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m$

$\text{Freezing point of pure solvent}=\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}$ ......(1)

where,

Freezing point of pure solvent = $-6.4^oC$

Freezing point of solution = $-13.6^oC$

i = Vant Hoff factor = 1 (for non-electrolytes)

$K_f$ = freezing point depression constant = $3.96^oC/m$

$m_{solute}$ = Given mass of solute (glycine) = ?

$M_{solute}$ = Molar mass of solute (glycine) = 75.07 g/mol

$w_{solvent}$ = Mass of solvent = 950. g

Putting values in equation 1, we get:

$-6.4-(-13.6)=1\times 3.96\times \frac{m_{solute}\times 1000}{75.07\times 950}\\\\m_{solute}=\frac{7.2\times 75.07\times 950}{1\times 3.96\times 1000}\\\\m_{solute}=129.66g=1.3\times 10^2g$

Hence, the mass of glycine that can be dissolved is $1.3\times 10^2g$

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Alchen [17]

The question is incomplete, here is the complete question:

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