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2 years ago

Calculate H for the process Hg2Cl2 ? 2Hg + Cl2 from the following information:

Chemistry

1 answer:

aleksandr82 [10.1K]2 years ago

6 0

Answer:

The enthalpy of the reaction of the reaction asked in the question is 265.2 kJ.

Explanation:

$Hg + Cl_2\rightarrow HgCl_2,\Delta H_1 = -224kJ$ ...(1)

$Hg + HgCl_2\rightarrow Hg_2Cl_2,\Delta H_2 = -41.2kJ$ ..(2)

$Hg_2Cl_2\rightarrow 2Hg + Cl_2,\Delta H_3=?$ ...(3)

(1)+(2)

$\Delta H_4=-224kJ+(-41.2kJ)= -265.2 kJ$

$2Hg+Cl_2\rightarrow Hg_2Cl_2,\Delta H_4=-265.2 kJ$ ...(4)

On reversing the equation (4)we get (3):

$\Delta H_4=-265.2 kJ =-(\Delta H_3)$

$Hg_2Cl_2\rightarrow 2Hg+Cl_2,\Delta H_3=265.2 kJ$

The enthalpy of the reaction of the reaction asked in the question is 265.2 kJ.

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The second-order diffraction for a gold crystal is at an angle of 22.20o for X-rays of 154 pm. What is the spacing between the c

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To calculate the spacing between the crystal planes, we use the equation given by Bragg, which is:

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where,

n = order of diffraction = 2

$\lambda$ = wavelength of the light = $154pm=1.54\times 10^{-10}m$ (Conversion factor: $1m=10^{12}pm$ )

d = spacing between the crystal planes = ?

$\theta$ = angle of diffraction = 22.20°

Putting values in above equation, we get:

$2\times 1.54\times 10^{-10}=2d\sin (22.20)\\\\d=\frac{2\times 1.54\times 10^{-10}}{2\times \sin (22.20)}\\\\d=4.07\times 10^{-10}m$

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Justification:

1) The depression of the freezing point of a solution is a colligative property, which means that it depends on the number of particles of solute dissolved.

2) The formula for the depression of freezing point is:

ΔTf = i * Kf * m

Where i is the van't Hoof factor which accounts for the dissociation of the solute.

Kf is the freezing molal constant and only depends on the solvent

m is the molality (molal concentration).

3) Since, you are assuming equal concentrations and complete dissociation of the given solutes, the solute with more ions in the molecular formula will result in the solution with higher depression of the freezing point (lower freezing point).

4) These are the dissociations of the given solutes:

a) NH4 Cl (s) --> NH4(+)(aq) + Cl(-) (aq) => 1 mol --> 2 moles

b) Co Br3 (s) --> Co(3+) (aq) + 3Br(-)(aq) => 1 mol --> 4 moles

c) K2SO4 (s) --> 2K(+) (aq) + SO4 (2-) (aq) => 1 mol --> 3 moles

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