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Chemistry 1 Marks Question

Thermodynamics · CBSE STD 11 Science (CBSE - English Medium). 61 questions.

Questions · Page 2 of 2

1 Marks Question

Question 511 Mark
Under what condition, the heat evolved or absorbed in a reaction is equal to its free energy change?
Answer
As $\Delta\text{G}=\Delta\text{H}-\text{T}\Delta\text{S}.$ Thus, $\Delta\text{G}=\Delta\text{H}$ only when either the reaction is carried out at 0K or the reaction is not accompanied by any entropy change, i.e. $\Delta\text{S}=0.$
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Question 521 Mark
Which have more entropy, real crystal or Ideal crystal and why?
Answer
Real crystal has more entropy because it has more disorderness or randomness.
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Question 531 Mark
Will the change in enthalpy of the system be zero in an adiabatic process?
Answer
Yes, In adiabatic process, $\Delta\text{H}=0$ because q = 0 and $\Delta\text{H}=\text{q}_\text{p}=0$
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Question 541 Mark
State a chemical reaction in which $\Delta\text{H}$ and $\Delta\text{U}$ are equal.
Answer
$\mathrm{H}_2(\mathrm{g})+\mathrm{I}_2(\mathrm{g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})$
Since $\Delta\text{n}=0,\ \therefore\Delta\text{H}=\Delta\text{U}.$
$\because$ Number of moles of gaseous reactants and products are equal.
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Question 551 Mark
When an ideal gas expands into vacuum, there is neither absorption nor evolution of heat. Why?
Answer
In an ideal gas, there are no intermolecular forces of attraction. Hence, no energy is required to overcome these forces. Moreover, when a gas expands against vacuum, work done is zero (because $\mathrm{Pe}_{\mathrm{xt}}=0$ ). Hence, internal energy of the system does not change, i.e. there is neither absorption nor evolution of heat.
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Question 561 Mark
Why does $\mathrm{NH}_4 \mathrm{NO}_3$ dissolve in water spontaneously even when this process is endothermic?
Answer
$\mathrm{NH}_4 \mathrm{NO}_3$ dissolves in water spontaneously even when this process is endothermic because entropy increasing due to free movement of ions on dissolving. $\Delta\text{S}=+\text{ve}$ favours the process and makes it spontaneous.
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Question 571 Mark
In the equation, $\text{N}_2(\text{g})+3\text{H}_2(\text{g})\rightleftharpoons2\text{NH}_3(\text{g})$ would be the sign of work done?
Answer
The sign of work done will be positive, i.e. work will be done on the system due to decrease in volume.
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Question 581 Mark
When 430J of work was done on a system, it lost 120J of energy as heat. Calculate the value of internal energy change $(\Delta\text{U})$ for this process.
Answer
$\Delta\text{U}=\text{q}+\text{w}$ $\text{q}=-120\text{J}$$\Delta\text{U}=-120+430$ $\text{w}=+430\text{J}$
$\Delta\text{U}=310\text{J}$
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Question 591 Mark
Which of the following has the highest entropy?
$\mathrm{H}_2(\mathrm{g}), \mathrm{F}_2(\mathrm{g}), \mathrm{Cl}_2(\mathrm{g}), \mathrm{Br}_2(\mathrm{g}), \mathrm{I}_2(\mathrm{g})$
Answer
$\mathrm{I}_2(\mathrm{g})$ has maximum entropy due to presence of largest number of electrons. has maximum entropy due to presence of largest number of electrons.
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Question 601 Mark
What are the ways by which the internal energy of a system can be changed?
Answer
  1. Exchanging heat with the surroundings.
  2. Work done on the system or by the system.
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Question 611 Mark
$\mathrm{CH}_4(\mathrm{~g})+2 \mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{I})$ ; $\Delta\text{H}=-890\text{kJ mol}^{-1}$
What is the calorific or fuel value of 1kg of $\mathrm{CH}_4$?
Answer
Calorific value/ kg $=\frac{890}{16}\times1000=55625\text{kJ/kg}$
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