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

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.

As q = 0, we have adiabatic process.

V₁ = 100L and V₂ = 800L

T₁ = 300K and T₂ = not given

For NH₃, γ = 4/3

Applying TV^γ = constant

(300)(100)^4/3-1 = (T)(800)^4/3-1

T = 300/2 = 150K

W= nR(T₂-T₁)/γ-1

= 1×8.314×150/(4/3-1)

= 3714 J

= 900 cal

w (reversible) < w (irreversible) (for compression process)

Work done is the area under the P−V curve. It can be seen in the curve above that the area under the P−V curve for irreversible compression of the gas is more than the area under the curve for reversible compression.

Thus, work done for irreversible compression is more than that for reversible compression.

C₄H₁₀(g) + 13/2O₂  →  4CO₂(g) + 5H₂O(l)

∆N_g = (n_g)_product - (n_g)_reactant

= 4-(13/2+1) = -7/2

or ∆N_g is negative.

∆H = ∆U + ∆N_gRT

SInce ∆n_g is negative therefore ∆H is less than ∆U.

Let atomic weight of x = M_x

atomic weight of y = M_y 

we know, 

mole = weight /atomic weight 

a/c to question, 

mole of xy₂ = 0.1 

so, 

0.1 = 10g/( M_x +2M_y) 

M_x + 2M_y = 100g -------(1)

for x₃y₂ ; mole of x₃y₂ = 0.05 

0.05 = 9/( 3M_x + 2M_y ) 

3M_x + 2M_y = 9/0.05 = 9 × 20 = 180 g ---(2) 

solve eqns (1) and (2)

2M_x = 80 

M_x = 40g/mol 

and M_y = 30g/mole

In endothermic reactions, the energy of reactants is less than that of the products. Potential energy diagram for endothermic reactions is,

Where Ea = activation energy of forwarding reaction

Ea' = activation energy of backwards reaction

ΔH = enthalpy of the reaction

From the above diagram,

Ea = Ea' + ΔH

Thus, Ea > ΔH

Δ_rH=Δ_rH_(product)-ΔrH_(reactant)

=3×-393+81+3×110-9.7

=-777.7kJ

ΔH = ΔU+Δn_gRT

AS Δn_g = 0, ΔH = ΔU

So, ΔU = -777.7 kJ