Chemistry Test 176

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Chemistry Test 176

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Weekly Quiz Competition

Question 1
4 / -1

From the following data, the heat of formation of Ca(OH)_2(s) at 18°C is ………..kcal:

A
-98.69

B
-235.43

C
194.91

D
98.69

Solution

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
Question 2
4 / -1

Which of the following statement(s) is/are correct :

Statement(a) : Δ_rH° for the reaction H₂O(g) → 2H(g) + O(g) is 925 kJ/mol

Statement(b) : Δ_rH° for the reaction OH(g) → H(g) + O(g) is 502 kJ/mol

Statement(c) : Enthalpy of formation of H(g) is-–218 kJ/mol

Statement(d) : Enthalpy of formation of OH(g) is 42 kJ/mol

A
Statement c

B
Statement a, b, d

C
Statement b,c

D
Statement a, d

Solution

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.
Question 3
4 / -1

A
a

B
b

C
c

D
d

Solution

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
Question 4
4 / -1

The pressure-volume work for an ideal gas can be calculated by using the expression The work can also be calculated from the pV– plot by using the area under the curve within the specified limits. When an ideal gas is compressed (a) reversibly or (b) irreversibly from volume Vi to Vf . choose the correct option.

A
w (reversible) < w (irreversible)

B
w (reversible) > w (irreversible)

C
w (reversible) = w (irreversible)

D
w (reversible) = w (irreversible) + p_ex.ΔV

Solution

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.
Question 5
4 / -1

During complete combustion of one mole of butane, 2658 kJ of heat is released. The thermochemical reaction for above change is Δ_fU⁰of formation of CH₄ (g) at certain temperature is –393 kJ mol⁻¹. The value of Δ_fH⁰is

A
equal to Δ_fU⁰

B
< Δ_fU⁰

C
zero

D
> Δ_fU⁰

Solution

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.
Question 6
4 / -1

Calculate the standard enthalpy of formation of CH₃OH(l) from the following data:

CH₃OH(I)+3/2O₂(g)→CO₂(g)+2H₂O(l); △_rH^θ=−726kJmol⁻¹

C_(graphite)+O₂(g)→CO₂(g); △_cH^θ=−393kJmol⁻¹

H₂(g)+1/2O₂(g)→H₂O(l); △_fH^θ=−286kJmol⁻¹

A
-339 kJ mol⁻¹

B
-239 kJ mol⁻¹

C
-269 kJ mol⁻¹

D
-209 kJ mol⁻¹

Solution
Question 7
4 / -1

Suppose the elements X and Y combine to form two compounds XY₂ and X₃Y₂. When 0.1 mole of XY₂ weighs 10 g and 0.05 mole of X₃Y₂ weighs 9 g, the atomic weights of X and Y are

A
40, 30

B
60, 40

C
20, 30

D
30, 20

Solution

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
Question 8
4 / -1

For an endothermic reaction when ΔH represents the enthalpy of the reaction in kJ mol^-1, the minimum value for the energy of activation will be

A
less than ΔH

B
Zero

C
More than ΔH

D
Equal to ΔH

Solution

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
Question 9
4 / -1

Enthalpies of formation of CO(g) , CO₂ (g) , N₂O (g) and N₂O₄ (g) are -110, - 393, 81 and 9.7 kJ mol^-1. Thus, Δ_rU for the reaction at 298 K is,

A
+777.7 KJ

B
- 777.7 KJ

C
-211.7 KJ

D
+211.7 KJ

Solution

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
Question 10
4 / -1

The equilibrium concentrations of the species in the reaction A + B ⇔ C + D are 2, 3, 10 and 6 mol L^-1, respectively at 300 K. ΔG⁰ for the reaction is (R = 2cal/mol K)

A
-137.26 cal

B
-1381.80 cal

C
-13.73 cal

D
1372.60 cal

Solution