Question 11 Mark
Why is standard enthalpy of formation of diamond not zero although it is an element?
AnswerIt is because diamond is not standard state of element. Standard state of carbon is graphite whose $\Delta_\text{f}\text{H}^\circ=0.$
View full question & answer→Question 21 Mark
Classify the following as extensive and intensive properties:
Molar heat capacity, Temperature, Enthalpy and Volume.
AnswerExtensive property: Volume, Enthalpy because these depend upon amount of substances.
Intensive property: Temperature, Molar heat capacity because these do not depend upon amount of substances.
View full question & answer→Question 31 Mark
If $\Delta\text{G}^\circ$ for reversible reaction is found to be zero, what is the value of its equilibrium constant?
Answer$\Delta\text{G}^\circ=0\ (\text{given})$$\Delta\text{G}^\circ=-2.303\text{ RT log K}=0$
$\Rightarrow\log\text{K}=0$
$\Rightarrow\log\text{K}=\log1$
$\Rightarrow\text{K}=1.$ $[\because\log1=0]$
View full question & answer→Question 41 Mark
The fact that enthalpy is a state function forms the basis of a very useful law. Name the law.
AnswerHess's law of heat summation.
View full question & answer→Question 51 Mark
Water can be lifted into the water tank at the top of the house with the help of a pump. Then why is it not considered to be spontaneous?
AnswerA spontaneous process should occur continuously by itself after initiation. But this is not so in the given case because water will go up as long as the pump is working.
View full question & answer→Question 61 Mark
For the same increase in volume, why work done is more if the gas is allowed to expand reversibly at higher temperature?
AnswerFor isothermal reversible expansion, $\text{W}=-\text{p}_\text{int}\times\Delta\text{V}.$
At higher temperature, internal pressure of the gas is more so work done is more.
View full question & answer→Question 71 Mark
Identify the species and physical state for which $\Delta_\text{f}\text{H}^\circ=0$ at 298K.
$\mathrm{Br}_2, \mathrm{Cl}_2, \mathrm{CH}_4$
Answer$\Delta_\text{f}\text{H}^\circ\text{Br}_2(\text{l})=0\ \ \Delta_\text{f}\text{H}^\circ\text{Cl}_2(\text{g})=0$
View full question & answer→Question 81 Mark
Give two examples of state functions.
Answer$\Delta\text{H}$ (Enthalpy change) and $\Delta\text{U}$ (Internal energy change) are state functions as they depend upon initial and final state and not on the path.
View full question & answer→Question 91 Mark
If $\Delta\text{H}$ for a reaction has a negative value, how would you know the sign requirement of $\Delta\text{S}$ for it so that the reaction is spontaneous at low temperatures?
Answer$\Delta\text{S}$ must be positive, then only $\Delta\text{G}$ will be negative and reaction will spontaneous.
$\because\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{T}\Delta\text{S}^\circ$
View full question & answer→Question 101 Mark
How can you say that universe is going towards chaos?
AnswerMost of the naturally occurring processes are accompanied by increase of randomness. Hence, randomness of the universe is continuously increasing. Thus, we are going towards chaos.
View full question & answer→Question 111 Mark
$\mathrm{H}_2(\mathrm{g})+\mathrm{Cl}_2(\mathrm{g}) \rightarrow 2 \mathrm{HCl}(\mathrm{g})+185 \mathrm{kJ}$. State whether this reaction Is exothermic or endothermic and why.
AnswerIt is an exothermic reaction because heat is being evolved.
View full question & answer→Question 121 Mark
Why for predicting the spontaneity of a reaction, free energy criteria is better than the entropy criteria?
AnswerCriteria of free energy change is better because it requires free energy change of the system only whereas the entropy change requires the total entropy change of the system and the surroundings.
View full question & answer→Question 131 Mark
At 1 atm will the $\Delta_{\mathrm{f}} \mathrm{H}^{\circ}$ be zero for $\mathrm{Cl}_2(\mathrm{g})$ and $\mathrm{Br}_2(\mathrm{g})$ ? Explain.
Answer$\Delta_{\mathrm{f}} \mathrm{H}^{\circ}$ for $\mathrm{Cl}_2(\mathrm{g})$ will be zero but $\Delta_{\mathrm{f}} \mathrm{H}^{\circ}$ for $\mathrm{Br}_2(\mathrm{g})$ will not be zero because liquid bromine is its elementary state and not gaseous bromine.
View full question & answer→Question 141 Mark
Write an expression in the form of chemical equation for the standard enthalpy of formation $(\Delta_\text{f}\text{H})$ of CO(g).
Answer$\text{C(s)}+\frac{1}{2}\text{O}_2(\text{g})\rightarrow\text{CO(g)}$1 mole of compound is funned from constituting elements in their standard states enthalpy change is $\Delta_\text{f}\text{H}^\circ.$
View full question & answer→Question 151 Mark
How is change in entropy during melting of solid related to its melting point?
Answer$\Delta\text{S}_\text{fusion}=\frac{\Delta\text{H}_\text{fusion}}{\text{Melting point in kelvin}}$
View full question & answer→Question 161 Mark
What happens to the internal energy of the system if:
- Work is done on the system?
- Work is done by the system?
Answer
- If work is done on the system, internal energy will increase.
$\because\Delta\text{U}=\text{q}+\text{w}$
- If work is done by the system, internal energy will decrease.
$\because\Delta\text{U}=\text{q}-\text{w}$
[work is -ve when done by the system.] View full question & answer→Question 171 Mark
Predict the sign of entropy change for the change.
$\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{I})$
Answer$\Delta\text{S}=-\text{ve}$ because disorder or randomness is less in liquid water than steam.
View full question & answer→Question 181 Mark
How is standard free energy change related to equilibrium constant?
Answer$\Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log \mathrm{K}$ where $\Delta \mathrm{G}^{\circ}$ is standard free energy change, R is gas constant, T is temperature in Kelvin, ' K ' is equilibrium constant.
View full question & answer→Question 191 Mark
Which of the following is an intensive property?
Surface tension, Mass, Volume, Enthalpy, Density.
AnswerSurface tension and density are intensive properties because these do not depend upon amount of substance.
View full question & answer→Question 201 Mark
Predict the sign of $\Delta\text{S}$ for the reaction:
$2 \mathrm{NaHCO}_3(\mathrm{s}) \rightarrow \mathrm{Na}_2 \mathrm{CO}_3(\mathrm{s})+\mathrm{CO}_2(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$
Answer$\Delta\text{S}=+\text{ve}.$ because more number of products are being formed and solid reactant is changing to solid and gas.
View full question & answer→MCQ 211 Mark
If the gibbs free energy is negative than reaction will be?
Answer$(D)$ : Always negative.
$\triangle G = 0, $ it means the reaction is equilibrium at standard conditions.
A negative value of $\triangle G,$ means spontaneous.
A positive value of $\triangle G,$ means non $-$ spontaneous.
View full question & answer→Question 221 Mark
Define standard enthalpy of formation.
AnswerIt is defined as enthalpy change when I mole of substance is fanned from constituenting elements in their standard states.
View full question & answer→Question 231 Mark
When is the entropy of a perfectly crystalline solid zero?
AnswerThe entropy of a perfectly crystalline solid is zero at absolute zero temperature, i.e. 0K or -273.15°C because there is perfect order and no disorder at this temperature.
View full question & answer→Question 241 Mark
Predict $\Delta\text{H}>\Delta\text{U}$ or $\Delta\text{H}$ is less than $\Delta\text{U}$ or $\Delta\text{H}=\Delta\text{U}.$
- C(graphite) $+\mathrm{O}_2(\mathrm{~g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})$
- $\mathrm{PCl}_5(\mathrm{~g}) \rightarrow \mathrm{PCl}_3(\mathrm{~g})+\mathrm{Cl}_2(\mathrm{~g})$
Answer
- $\Delta\text{H}=\Delta\text{U}$ because $\Delta\text{n}=0$ i.e. number of moles of gaseous reactants and products are equal.
- $\Delta\text{H}>\Delta\text{U}$ because $\Delta\text{n}=1\ [\Delta\text{H}=\Delta\text{U}+\Delta\text{nRT}]$
$\because$ Numbers of moles of gaseous products are more than gaseous reactants. View full question & answer→Question 251 Mark
What is the limitation of first law of thermodynamics?
AnswerIt cannot tell us the direction of the process i.e. feasibility of Process.
View full question & answer→Question 261 Mark
Determine the sign of entropy change in $\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})$.
Answer$\Delta\text{S}=+\text{ve}$ because it is endothermic process and spontaneous. $\Delta\text{G}^\circ$ will -ve only if $\Delta\text{S}$ is +ve.$\because\Delta\text{G}^\circ=\Delta\text{H}^\circ-\text{TDS}^\circ$
View full question & answer→Question 271 Mark
Why does entropy increase on mixing of two gases?
AnswerThe disorder increases when two gases are mixed together, that is why entropy increases.
View full question & answer→Question 281 Mark
Give an example of a spontaneous process which is endothermic.
Answer$\mathrm{N}_2(\mathrm{g})+\mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{NO}(\mathrm{g})$
It is endothermic as well as spontaneous because entropy is increasing and $\Delta\text{G}=-\text{ve.}$
View full question & answer→MCQ 291 Mark
Choose the correct answer. The enthalpies of all elements in their standard states are :
- A
- ✓
- C
$ < 0$
- D
Different for each element.
AnswerThe enthalpy of all elements in their standard state is zero.
View full question & answer→Question 301 Mark
Which one of the following is not extensive state function?
Enthalpy change, Internal energy change and Pressure.
AnswerPressure is not a state function because it depends upon path.
View full question & answer→Question 311 Mark
Neither q nor W is a state function but q + W is state function. Explain why?
AnswerBoth q and W are not state functions but q + W is equal to $\Delta\text{U}$ which is a state function.
View full question & answer→Question 321 Mark
Why is entropy of a solution higher than that of pure liquid?
AnswerIn solution, there is more disorderness than in pure liquid, so entropy is more.
View full question & answer→Question 331 Mark
What is the effect of temperature on entropy?
AnswerEntropy increases with increase in temperature because disorder or randomness increases.
View full question & answer→Question 341 Mark
One mole of acetone requires less heat to vapourise than 1mol of water. Which of the two liquids has higher enthalpy of vapourisation?
AnswerWater has higher enthalpy of vapourisation. $(\Delta\text{H}_\text{r})_\text{water}>(\Delta\text{H}_\text{r})_\text{acetone}$
View full question & answer→Question 351 Mark
Name the thermodynamic system to which following belong:
- Human body.
- Milk in thermos flask.
Answer
- Human body is an open system because it can exchange matter as well as energy with the surroundings.
- Milk in thermos flask is an isolated system which can neither exchange matter nor energy with the surroundings.
View full question & answer→Question 361 Mark
The standard molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{I})$ is $70 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. Will the standard molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{s})$ be more, or less than $70 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$ ?
AnswerThe standard molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{I})$ is $70 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$. The solid form of $\mathrm{H}_2 \mathrm{O}$ is ice. In ice, molecules of $\mathrm{H}_2 \mathrm{O}$ are less random than in liquid water. Thus, molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{s})<$ molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$. The standard molar entropy of $\mathrm{H}_2 \mathrm{O}(\mathrm{s})$ is less than $70 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}$.
View full question & answer→Question 371 Mark
State why heat changes in physical and chemical processes are indicated by enthalpy changes and not by internal energy changes.
AnswerMost of the processes are carried out in open container, at constant pressure. Therefore, enthalpy changes indicate heat changes in processes and not internal energy changes. Enthalpy changes are measured at constant pressure where as internal energy change is measured at constant volume.
View full question & answer→MCQ 381 Mark
In the following Questions, the Assertion and Reason have been put forward. Read the statements carefully and choose the correct alternative from the following:
Assertion : Heat capacity is the amount of heat required to raise the temperature of a body by $1K.$
Reason : Heat capacity is an extensive property and It depends upon the size of the body.
- A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
- ✓
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
- C
Assertion is true but Reason is false.
- D
Both Assertion and Reason are false.
AnswerCorrect option: B. Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
View full question & answer→Question 391 Mark
Why in some reactions heat is evolved while some reactions take place only on absorption of heat?
AnswerEvery substance has energy stored in it in the form of heat content. If heat content of reactants $\left(H_R\right)$ is greater than that of products $\left(H_p\right)$, heat is evolved. If $H_R<H_p$, heat is absorbed.
View full question & answer→Question 401 Mark
What is the value of internal energy for 1 mole of a monoatomic gas?
Answer$\text{U}=\frac{3}{2}\text{RT}$ for monatomic gas.
View full question & answer→Question 411 Mark
Name the state variables which remain constant in:
- Isobaric process.
- Isothermal process.
Answer
- Isobaric process: In this process, pressure remains constant i.e. $\Delta\text{p}=0.$
- Isothermal process: In this process, temperature remains constant i.e. $\Delta\text{T}=0.$
View full question & answer→Question 421 Mark
Predict the sign of $\Delta\text{S}^\circ$ for the following reaction:
$2 \mathrm{H}_2 \mathrm{S}(\mathrm{g})+3 \mathrm{O}_2(\mathrm{g}) \rightarrow 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})+2 \mathrm{SO}_2(\mathrm{g})$
Answer$\Delta\text{S}^\circ,$ i.e. entropy decreases during the reaction because 5 moles of gaseous reactants change into 4 moles of gaseous products. Hence, the sign of $\Delta\text{S}^\circ$ will be negative in the given reaction.
View full question & answer→MCQ 431 Mark
In the following Questions, the Assertion and Reason have been put forward. Read the statements carefully and choose the correct alternative from the following:
Assertion : An isolated system is the one which can neither exchange matter nor energy with the surroundings.
Reason : It should be noted that every system is perfectly isolated.
- A
Both Assertion and Reason are true and Reason is the correct explanation of Assertion.
- B
Both Assertion and Reason are true but Reason is not the correct explanation of Assertion.
- ✓
Assertion is true but Reason is false.
- D
Both Assertion and Reason are false.
AnswerCorrect option: C. Assertion is true but Reason is false.
View full question & answer→Question 441 Mark
For the reaction, $\mathrm{N}_2(\mathrm{g})+3 \mathrm{H}_2(\mathrm{g}) \rightarrow 2 \mathrm{NH}_3(\mathrm{g})$, predict whether the work is done on the system or by the system.
AnswerVolume is decreasing, therefore, work is done on the system.
$\because\ \text{w}=-\text{P}\Delta\text{V,}$
$\Delta\text{V}=-\text{ve}$
$\text{w}=+\text{ve}$
View full question & answer→Question 451 Mark
The energy released in the neutralisation of $\mathrm{H}_2 \mathrm{SO}_4$ and KOH is $57.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$. Therefore, calculate the value of $\Delta \mathrm{H}$ for the reaction:
$\mathrm{H}_2 \mathrm{SO}_4+2 \mathrm{KOH} \rightarrow \mathrm{~K}_2 \mathrm{SO}_4+2 \mathrm{H}_2 \mathrm{O}$
Answer$\Delta \mathrm{H}=-2 \times 57.1=-114.2 \mathrm{~kJ}$ because 2 moles of $\mathrm{H}^{+}$from $\mathrm{H}_2 \mathrm{SO}_4$, are reacting with 2 moles of $\mathrm{OH}^{-}$ from 2 KOH to form 2 moles of water.
View full question & answer→Question 461 Mark
If enthalpy of fusion and enthalpy of vaporisation of sodium metals are 2.6 and $98.2 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively, what is the enthalpy of sublimation of sodium.
Answer$\Delta_\text{sub}\text{H}^\circ=\Delta_\text{fus}\text{H}^\circ+\Delta_\text{vap}\text{H}^\circ$
$=2.6+98.2=100.8\text{kJ mol}^{-1}$
View full question & answer→Question 471 Mark
Air contains about $99 \%$ of $\mathrm{N}_2$ and $\mathrm{O}_2$ gases. Why do not they combine to form NO under the standard conditions? Standard Gibbs energy of formation of $\mathrm{NO}(\mathrm{g})$ is $86.7 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
AnswerSol. For the combination of N , and $\mathrm{O}_2$ to form NO , the standard Gibbs energy of formation, is + ve
$\frac{1}{2} \mathrm{~N}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{NO}(\mathrm{~g})\left(\Delta_{\mathrm{f}} \mathrm{G}_{\mathrm{NO}}^{\circ}=+86.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$
Therefore, this reaction is non-spontaneous under the standard conditions and hence $\mathrm{N}_2$ and $\mathrm{O}_2$ do not combine.
View full question & answer→Question 481 Mark
Identify the state functions and path functions out of the following : enthalpy, entropy, heat, temperature, work, free energy.
AnswerState functions: Enthalpy, entropy, temperature, free energy Path functions: Heat, work.
View full question & answer→Question 491 Mark
Under what conditions $\Delta\text{H}$ and $\Delta\text{U}$ are equal?
AnswerWhen $\Delta\text{n}=0,\ \Delta\text{H}=\Delta\text{U}$
View full question & answer→Question 501 Mark
What is the value of $\Delta\text{G}$ when ice and water are in equilibrium?
Answer$\Delta\text{G}=0$ at equilibrium.
View full question & answer→