Thermochemistry - Free MCQ Practice Test with solutions, NEET

∆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%.

Trans-2-butene + 6O₂(g) → 4CO₂(g) + 4H₂O(l)
∆H = -647.0 kcal/mol ---(I)
1 - butene + 6O₂(g) → 4CO₂(g) + 4H₂O(l)
ΔH = -649.8 kcal/mol ---(II)
On (I)-(II)
Trans-2-butene → 1 - butene
ΔH = 2.7 kcal/mol
ΔH = (H_f)_1-Butene - (H_f)_{trans-2-butene}
= H₂ - H₁
H₂ - H₁ = 2.7 -----(A)
And 9ΔH₁ + 5ΔH₂ = 0 -----(B)
On solving eqn (A) and (B), we get
H₁ = -1.0 and H₂ = 1.8

During bond breakage energy is absorbed and during bond formation it is released. From the reaction we can say that 1 C-H bond is broken 1 Cl-Cl bond is broken 1 c-cl bond is formed and 1 h-cl bond is formed. so using the sign conventions the equation becomes
x+y-84-103= -25 (∆H = -25)
5x=9y ..putting x=9/5y we get y = 57.75 kCal

Let's check statement (a) 
Statement(a) : ΔrH° for the reaction H₂O(g) → 2H(g) + O(g) is 925 kJ/mol
For this, we need - (II) + (III) + ½ (IV)
We get, H₂O(g) → 2H(g) + O(g) - (-242)+436+½ 495 = 925.5 kJ mol^-1
So it is true.
Let's check statement (b)
Statement(b) : ΔrH° for the reaction OH(g) → H(g) + O(g) is 502 kJ/mol
For this we need -(I)+½ (III)+½ (IV)
We get OH(g)   →   H(g)  + O(g)       -(42) + ½ (436) + ½ (495) = 423.5 kJ mol-1
So statement (b) is wrong.
Let's check statement (c)
Statement(c) : Enthalpy of formation of H(g) is -218 kJ/mol
We can see that for enthalpy of formation, we need to divide eqn (III) by 2
So, it would become :- 
½ H₂(g) → H(g)
436/2 = 218 kJ
So, statement (c) is wrong.
Let's check statement (d)
Statement(d) : Enthalpy of formation of OH(g) is 42 kJ/mol
For that, we have eqn (I) as it is. So, statement (d) is correct.

∆H = ∆E + ∆n_gRT
176 = ∆E + 1×8.314×1240/1000
∆E = 176 - 10.30 = 165.69 kJ

10⁷ erg = 1 joule and 4.2 joules = 1 cal
Therefore 1 cal > 1 joule > 1 erg

The enthalpies of these reactions are less than zero, and are therefore exothermic reactions. A system of reactants that absorbs heat from the surroundings in an endothermic reaction has a positive \(ΔH\), because the enthalpy of the products is higher than the enthalpy of the reactants of the system.

Mol. wt. of methane =16gm
Heat liberated during the combustion of 4gm methane = 2.5kcal
Heat liberated during the combustion of 16gm methane = 2.5/4×16=10kcal
Hence the heat of combustion of methane is 10 kcal.

NH₄NO₃ (s) → N₂O (g) + 2H₂O ;     ΔH = -37.0 KJ/mol from the above data, we can show that ΔH is for 1 mole of NH₄NO₃.
For 2.5 g of NH₄NO₃, we have 2.5/80 = 0.03125 moles
Therefore, heat released = (-37)×(0.03125) = -1.16 kJ

The correct answer is Option B.

Ca_(s) + O2_(g) + H2_(g) → Ca(OH)₂ , ΔHf = ?

Desired equation = eq (iii) + eq(i) - eq (ii)

ΔH_f = (−151.80)+(−15.26)−(−68.37)
ΔH_f = (-151.80)+(-15.26)-(-68.37)
ΔH_f = −235.43KCalmol⁻¹

This is not fixed. It depends from compound to compound. 
A negative enthalpy of formation indicates that the formation of a compound is exothermic---the amount of energy it takes to break bonds is less than the amount of energy that is released when making the bonds.
A positive enthalpy of formation indicates that the formation of a compound is endothermic---the amount of energy it takes to break bonds is greater than the amount of energy that is released when making the bonds.

The reason behind is that combustion reactions are always exothermic. So the enthalpy of combustion is always less than zero.

HA + OH⁻ → H₂O + A⁻ + q1 -----(1)
 H₂O + q2 → H⁺ + OH⁻ -----(2)
Adding eqn 1 and 2,
HA + q2 → H⁺ + A⁻  + q1
Hence , bond dissociation energy of HA is (q2 - q1)

BaCl₂(s) + aq. ⟶ BaCl_{2 aq.}
∆H = -a kJ ------(i)
BaCl₂ 2H₂O(s) + aq. ⟶ BaCl_{2 aq.} + 2H₂O
∆H = b kJ -----(ii)      
Enthalpy of hydration means that the molecule is just hydeedratt
The reaction should be 
(i) - (ii) or
BaCl₂(s) + 2H₂O ⟶ BaCl₂.2H₂O       ∆H = -a - b kJ

Let heat evolved in the 1st case is Q₁ and that in the 2nd case is Q₂. Then Q₂ = 1/2Q₁
However, Q₁ = 1000T₁ and Q₂ = 500T₂
Therefore, 500T₂ = ½ 1000T₁ i.e. T₁ = T₂

According to Hess law, the thermal effects of a reaction depends upon the initial and final conditions of the reacting substances. It does not depend upon the immediate steps.

According to Hess law, the thermal effects of a reaction depends upon the initial and final conditions of the reacting substances. It does not depend upon the immediate steps.

For the reaction of 1 mole of H⁺ and OH^-,we have 13.6 kcal energy released. In H₂SO₄, we have 2 moles of H⁺. So for its complete neutralisation, we need 2 moles of NaOH. 
So in the end, 2 moles of H⁺ reacts with2 moles of OH^- and 13.6 2 = 27.4 kcal energy is released.