Test: Thermochemistry (June 13) - JEE MCQ

∆H = ∆U+∆ngRT
For ∆H not equal ∆U, ∆ng should not equal zero. 
This happens only in option d, where ∆ng  = -2
 

In equation d, H⁺(aq) is formed from H₂(g). Enthalpy of formation of both entity is considered to be zero. So, ∆rH (∆H_product - ∆H_reactant) is zero.

Reaction is CaO + C → CaC₂+CO
1.28 kg of CaC₂ means 20 mole CaC₂ is needed
For 1 mole dH=-26-14-(-152)=112
For 20 Mole 112×20=2240

No. of moles of HCl = 5 millimoles
No. of moles of NaOH = 5 millimoles
Mass of solution mixed = 50 gm+50 gm=100 gm
ΔH=−cmΔT(c=4.18 kJKg⁻¹)
⇒ ΔH=−4.18×0.1×3
⇒ ΔH=−1.254 kJ (For 5 millimoles of water formed)
For 1 mole water = −1.254/5×10⁻³
=−2.5×10²kJ

The correct answer is option C
Ideally, the enthalpy of neutralization should be - 57.3 K.J + 1.5 K.J = - 55.8 KJ.
But it is - 56.1 KJ.
∴ Energy used for neutralization 
= 57.3 - 56.1 = 1.2 KJ
∴ Percent ionization of weak acid

∴ % of weak acid in solution = 20%

∆H given in the question is for one mole of C (g). If 6 gm of diamond and graphite are burnt in oxygen then the C (diamond) will first convert to graphite and then it will form CO₂. While C (graphite) will directly form CO₂. So due to the conversion of diamond into graphite, we will get extra heat. And since we have 6gm (0.5 mol) of diamond, so the heat released will be 0.5×1.9 kJ or 0.95 kJ more than the second case.

Combustion of hydrogen:
H₂ + ½ O₂→H₂O (H−O−H)
As water contains two O−H bonds.
So, combustion enthalpy of hydrogen is:
ΔH= Bond energy of reactant − Bond energy of the product − Enthalpy of vaporization. 
ΔH=x₁+2₂−2x₃−x₄

The formation of NCl₃ be like
½ N₂ + 3/2Cl₂ ⇋ NCl₃
We can see that for this setup, we need to have eqn (ii) divided by 2, reversing eqn (iii) and multiplying it by 3/2 and then adding all these to equation (i).
So option a is correct.

-57 kJ of heat evolved when 1 mole of NaOH reacted with an acid.
We have 1 mole of H⁺ and 0.75 moles of OH^-. So OH^- iis limiting reagent. Or only 0.75 moles of OH^- will be used. So for 1 mole of OH^-, we have  -57 kJ heat released. Therefore for 0.75 moles, we have ¾× -57 kJ heat released.

Au(OH)₃ + 4HCl → HAuCl₄+ 3H₂O…(1) ∆H₁=-28kcal
Au(OH)₃ + 4HBr → HAuBr₄ + 3H₂O ...(2) ∆H₂= -36.8kcal
To convert HAuBr₄ to HAuCl₄, the net reaction is
HAuBr₄ + 4HCl→ HAuCl₄ + 4HBr ...∆H=?
For the above reaction: 
∆H =∆H₁ - ∆H₂
      = -28 - (-36.8) = 8.8 kcal
Thus to convert one mole of HAuBr₄ to HAuCl₄ we require 8.8 kcal energy but since the energy absorbed is 0.44 kcal.
hence %conversion =[(0.44)/(8.8)] x 100 = 5%
Hence the percentage conversion is 5%.