Answer:
ΔH of the reaction is -802.3kJ.
Explanation:
Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.
Using the reactions:
<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ
<em>(2) </em>Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k
J
<em>(3) </em>H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ
The sum of (2) - (1) produce:
CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ
And the sum of this reaction with 2×(3) produce:
CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =
<em>-802.3kJ</em>
<span>PV/T = P'V'/T'
660 x 1.00/295.2 = P' x 1.00/317.8
P'=710.5 torr</span>
Here, we should use combined gas law which can be derived from combined gas law, “PV=nRT”. Rearranging, we can get PV/T=nR. Then we can set the two states in the problem together to get
P1V1/T1 = P2V2/T2
Then just plug in and solve algebraically.
Hope this helps
Answer : The temperature of liquid is, 369.9 K
Explanation :
The Clausius- Clapeyron equation is :
where,
= vapor pressure of liquid at 373 K = 681 torr
= vapor pressure of liquid at normal boiling point = 760 torr
= temperature of liquid = ?
= normal boiling point of liquid = 373 K
= heat of vaporization = 40.7 kJ/mole = 40700 J/mole
R = universal constant = 8.314 J/K.mole
Now put all the given values in the above formula, we get:
Hence, the temperature of liquid is, 369.9 K
The correct answer is less wind and more wind. Good luck :)
