Amount of heat evolved combustion of ethane is 900 kJ/mol. At 300 k and 1 atm. C2H6(g) + 7/2 O2(g) - 2 Co2(g) + 3H2O(l). Calculate W, ∆H, ∆U for combustion of 12×10^-3 kg ethane.
Answers
Given reaction :
- C₂H₆ (g) + 7/2 O₂ (g) → 2 CO₂(g) + 3H₂O(l)
- ΔH꜀ₒₘ = 900 KJ / mol
- T = 300 K
- Mass of ethane = 12 x 10³ kg
Work done is given by,
- W = Δng RT
Here,
Δn g = stoichiometric coefficients of gaseous products - stoichiometric coefficients of gaseous reactants
Δn g = 2 - (7/2 + 1) = 2 - 9/2 = -5/2
Hence,
⇒ W = - 5/2 x 8.314 x 300
⇒ W = - 5 x 8.314 x 150
⇒ W = - 6235.5 J
Now, let us find the moles of ethane,
Mass = 12 x 10-³ kg
Molar mass = 2(C) + 6(H) = 24 + 6 = 30 g/mol
⇒ moles = mass/molar mass
⇒ moles = 0.012/30
⇒ moles = 0.4 moles
Given condition,
ΔH of combustion for 1 mole of ethane is - 900 KJ/mol
So, ΔH of 0.4 moles of ethane = 0.4 x - 900 = -360 KJ/mol
ΔH = -360 KJ / mol
We know that, as per first law of thermodynamics,
ΔH = ΔU + W
⇒ ΔU = ΔH - W
⇒ ΔU = -360000 - (-6235.5 )
⇒ ΔU = 353764 J
⇒ ΔU ≈ 354 KJ
Answer:
Answer :
Iron. The blue colour of aqueous copper sulphate solution can be changed to place green by immersing an iron rod in it.
Explanation :
Out of the given options, all except silver can displace copper from its solution. But, we are also asked that the solution should be changed to a colour of pale-green. Hence, the answer is iron beacuse the colour of iron sulphate solution is pale-green.
When an iron rod is immersed in the copper sulphate solution it displaces copper from its solution and forms iron sulphate solution which gives a pale green colour to the resultant solution.
Reaction :
\sf{CuSO_4(aq)+Fe(s)\longrightarrow{}\underset{\textsf{pale green}}{FeSO_4(aq)}+Cu(s)}CuSO4(aq)+Fe(s)⟶pale greenFeSO4(aq)+Cu(s)
_____________
Additional Information :
Displacement reaction :
A type of chemical reaction in which an atom or group of atoms displace another atom or group of atoms of a compound. Displacement reactions are also called single-displacement reactions.
Examples :
\sf{CuCl_2(aq)+Zn(s)\longrightarrow{}ZnCl_2(aq)+Cu(s)}CuCl2(aq)+Zn(s)⟶ZnCl2(aq)+Cu(s)
\sf{CuSO_4(aq)+Zn(s)\longrightarrow{}ZnSO_4(aq)+Cu(s)}CuSO4(aq)+Zn(s)⟶ZnSO4(aq)+Cu(s)
\sf{CuCl_2(aq)+Pb(s)\longrightarrow{}PbCl_2(aq)+Cu(s)}CuCl2(aq)+Pb(s)⟶PbCl2(aq)+Cu(