Chemistry NCERT Exemplar Solutions Class 11th Chapter Six

This is a Multiple Choice Questions as classified in NCERT Exemplar

The correct option is B $°$
The balanced equation for the combustion of methane is:

CH₄ (g)+2O₂ (g)→CO₂ (g)+2H₂O (l)

Here,  Δng=1−3=−2

ΔH $°$ =ΔU $°$ +ΔngRT

ΔH $°$ =−393−2RT

∴ΔH $°$ $°$

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (iii)

Standard enthalpy of combustion is defined as the enthalpy change per mole (or per unit amount) of a substance when it undergoes combustion and all the reactants and products being in their standard states at the specified temperature.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

Specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or 1 Kelvin). That is why it is an intensive property which does not depend on mass.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (i)

Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

The presence of reactants in a closed vessel made of conducting material e.g., copper or steel is an example of a closed system.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Thermodynamics is not concerned about how and at what rate these energy transformations are carried out but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state.

This is a Short Answer Type Questions as classified in NCERT Exemplar

We know that the amount of work done =-p_ext? V

On substituting the values in the formula, we get,

-2bar* (50-10)L=-80Lbar

According to the described problem,1 LBar = 100J

Therefore, -80 L bar= (-80*100)= -8000J

= -8kJ, which is the amount of work done

The significance of the negative sign states that the work is done on the surroundings of the system. In the case of reversible expansion, the work done will be more.

This is a Short Answer Type Questions as classified in NCERT Exemplar

We can conclude from the figure that this change is a reversible change.

Now,

W= -2.303nRT log $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

But,  p₁V₁ = p₂V_{2 = $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$} = $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$ = $\frac{2}{1}$ =2

 W= -2.303nRT log $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$

 = -2.303 *8.314*1*298*log2

= -2.303 *8.314*298*0.3010J

= -1717.46J

This is a Short Answer Type Questions as classified in NCERT Exemplar

No, for the state of spontaneity, the enthalpy change is not the only criteria. Entropy also needs to be taken into account here.

This is a Short Answer Type Questions as classified in NCERT Exemplar

Throwing a stone from ground to roof

b) the reaction involved is a process where the energy decreases after the reaction. It can be represented as:In process b), potential energy/enthalpy change is a contributing factor to the spontaneity.