Chemistry, 21.08.2019 05:00 jessicasbss6840

[co(h2o)6]2+(aq) + 4cl-(aq) ⇌ [cocl4]2-(aq) + 6h2o(l)
concentrations at 25 degrees c
h+ 9.84368e-8
oh- 9.84368e-8
co (h2o)6^2 0.966539
cl- 1.86616
cocl4^2- 0.0334612
concentrations at 65 degree c
h+ 3.73844e-7
oh- 3.73844e-7
co (h2o)6^2 0.765375
cl- 1.06150
cocl4^2- 0.234625
a. write a k expression for this reaction (note that that liquid water shouldn’t be included in the k expression).
b. use the k expression and equilibrium concentrations on the left to determine the k value at 25 deg c. show all work for full credits.
c. what is the equilibrium constant k’ 65 degrees c?
d. compare the k’ constants; are the value difference agree or against with endo or exothermic determination?

Answers: 2

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Answers

Answer from: thebasedgodchri

For A: The expression for $K_{eq}$ is given below.

For B: The value of $K_{eq}$ at 25°C is 0.0185512

For C: The value of $K_{eq}$ at 65°C is 0.2887886

For D: The reaction is endothermic in nature.

Explanation:

For A:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_{eq}$

For the given chemical reaction:

$[Co(H_2O)_6]^{2+}(aq.)+4Cl^-(aq.)\rightleftharpoons [CoCl_4]^{2-}(aq.)+6H_2O(l)$

The expression of $K_{eq}$ for above equation without the concentration of liquid water is:

$K_{eq}=\frac{[CoCl_4]^{2-}}{[Co(H_2O)_6]^{2+}[Cl^-]^4}$ ......(1)

The expression is written above.

For B:

We are given:

$[CoCl_4]^{2-}=0.0334612M$

$[Co(H_2O)_6]^{2+}=0.966539M$

[Cl^-]=1.86616M

Putting values in equation 1, we get:

$K_{eq}=\frac{0.0334612}{0.966539\times 1.86616}=0.0185512$

Hence, the value of $K_{eq}$ at 25°C is 0.0185512

For C:

We are given:

$[CoCl_4]^{2-}=0.234625M$

$[Co(H_2O)_6]^{2+}=0.765375M$

[Cl^-]=1.06150M

Putting values in equation 1, we get:

$K_{eq}=\frac{0.234625}{0.765375\times 1.06150}=0.2887886$

Hence, the value of $K_{eq}$ at 65°C is 0.2887886

For D:

For Endothermic reactions, $\Delta H0$ , which is positive

For Exothermic reactions, $\Delta H$ , which is negative

To calculate $\Delta H$ of the reaction, we use Van't Hoff's equation, which is:

$\ln(\frac{K_{65^oC}}{K_{25^oC}})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_{65^oC}$ = equilibrium constant at 65°C = 0.2887886

$K_{25^oC}$ = equilibrium constant at 25°C = 0.0185512

$\Delta H$ = Enthalpy change of the reaction = ?

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 25^oC=[25+2730]K=298K

T_2 = final temperature = 65^oC=[65+2730]K=338K

Putting values in above equation, we get:

$\ln(\frac{0.2887886}{0.0185512})=\frac{\Delta H}{8.314J/mol.K}[\frac{1}{298}-\frac{1}{338}]\\\\\Delta H=57471.26J/mol$

As, the calculated value of $\Delta H0$ . Thus, the reaction is endothermic in nature.

Answer from: Quest

i just learnt this recently but i need to look through my notes again

Answer from: Quest

answer is: naturally occurring solid, inorganic compound or element is mineral.

minerals usually have crystalline form, that means than crystal constituents (atoms, molecules or ions) are arranged in a highly ordered microscopic structure, forming a crystal lattice that extends in all directions.

for example magnetite is a mineral of iron oxide (fe₂o₃).

calcite is a carbonate mineral and the most stable polymorph of calcium carbonate (caco₃).

dolomite is an carbonate mineral composed of calcium magnesium carbonate (camg(co₃)₂).