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Gnesinka [82]
3 years ago
13

A solution is 40.0% by volume benzene (C6H6) in carbon tetrachloride at 20C. The vapor pressure of pure benzene at this temperat

ure is 74.61 mmHg and its density is 0.87865 g/cm3; the vapor pressure of pure carbon tetrachloride is 91.32 mmHg and its density is 1.5940 g/cm3. If this solution is ideal, its total vapor pressure at 20C is
Chemistry
1 answer:
Alexandra [31]3 years ago
6 0

Answer:

84.30 mm Hg

Explanation:

In 100 cm³ of solution we have: 40 cm³ C6H6 and 60 cm³ CCl4. Given the densities we can calculate their masses and number of moles, and since by Raoult´s law

Ptotal = XAPºA + XBPºB

where XA= mol fraction =na/(na +nb) and PºA vapor pressure pure of pure component A

m C6H6 = 40 cm³ x 0.87865 g/cm³ = 35.146 g

mol C6H6 = 35.146 g/ 78.11 g/mol = 0.45 mol

mass CCl4 = 60 cm³ x 1.5940 g/cm³ = 95.640 g

mol CCl4 = 95.640 g / 153.82 g/mol = 0.62 mol

mol tot = 1.07

XC6H6 = 0.45/ 1.07 = 0.42      XCCl4 = 0.62/1.07 =0.58

Ptot (mmHg) = 0.42 x 74.61 + .58 x 91.32 = 84.30 mmHg

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Determine the ΔH for the following reaction 2NH3 + 5/2O2 = 2NO(g) + 3 H2O(g)
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The enthalpy change, ΔH for the following reaction 2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g) is -452.76 kJ.

<h3>What is enthalpy change, ΔH, of a reaction?</h3>

The enthalpy change of a reaction is the heat changes that occurs when a reaction proceeds to formation of products.

  • Enthalpy change, ΔH = ΔH of products - ΔH of reactants

The equation of the reaction is given below

2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g)

\Delta{H_{f}\:of\:NO = 90.25 kJ; \Delta{H_{f}\:of\:H_{2}O =-241.82kJ;  \Delta{H_{f}\:of\:NH_{3} =-46.1 kJ;  \Delta{H_{f}\:of\:O_{2} =0

\Delta{H_{f}\:of\:rxn = (90.25*2)+(-241.82*3)-( -46.1*2)= -452.76\:kJ

Therefore, the enthalpy change, ΔH for the following reaction 2\:NH_{3}(g) + \frac{5}{2}\:O_{2}(g)\rightarrow 2\:NO (g) + 3\:H_{2}O (g) is -452.76 kJ.

Learn more about enthalpy change at: brainly.com/question/14047927

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2 years ago
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