Why is carbon dioxide nonpolar?
1 answer:
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Answer:
a. -206,4kJ
b. Surroundings will gain heat.
c. -115kJ are given off.
Explanation:
It is possible to obtain ΔH of a reaction using Hess's law that consist in sum the different ΔH's of other reactions until obtain the reaction you need.
Using:
<em>(1) </em>C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ
<em>(2) </em>H₂(g) + O₂(g) → H₂O₂ (l) ΔH: -187.8 kJ
<em>(3) </em>H₂(g) + 1/2O₂(g) → H₂O(l) ΔH: -285.8 kJ
It is possible to obtain:
C₆H₄(OH)₂(l) + H₂O₂(l) → C₆H₄O₂(l) + 2H₂O(l)
From (1)-(2)+2×(3). That is:
<em>(1) </em>C₆H₄(OH)₂(l) → C₆H₄O₂(l) + H₂(g) ΔH: +177.4 kJ
<em>-(2) </em>H₂O₂(l) → H₂(g) + O₂(g) ΔH: +187.8 kJ
<em>2x(3) </em>2H₂(g) + O₂(g) → 2H₂O(l) ΔH: 2×-285.8 kJ
The ΔH you obtain is:
+177,4kJ + 187,8kJ - 2×285.8 kJ =<em> -206,4kJ</em>
b. When ΔH of a reaction is <0, the reaction is exothermic, that means that the reaction produce heat and the <em>surroundings will gain this heat.</em>
c. 20,0g of H₂O are:
20,0g× = <em>1,11 mol H₂O</em>
As 2 moles of H₂O are produced when -206,4kJ are given off, when 1,11mol of H₂O are produced, there are given off:
1,11mol H₂O× =<em> -115kJ</em>
I hope it helps!
Answer:
28.7664 kJ /mol
Explanation:
The expression for Clausius-Clapeyron Equation is shown below as:
Where,
P is the vapor pressure
ΔHvap is the Enthalpy of Vaporization
R is the gas constant (8.314×10⁻³ kJ /mol K)
c is the constant.
The graph of ln P and 1/T gives a slope of - ΔHvap/ R and intercept of c.
Given :
Slope = -3.46×10³ K
So,
- ΔHvap/ R = -3.46×10³ K
<u>ΔHvap = 3.46×10³ K × 8.314×10⁻³ kJ /mol K = 28.7664 kJ /mol</u>
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