Online Test: Thermodynamics | Class 11 Chemistry

Question 1 of 20

The first law of thermodynamics is:

$\Delta U=q-w$

$\Delta U=q+w$

$q=\Delta U-w$

$\Delta U=w-q$

Explanation: With the IUPAC convention $\Delta U=q+w$, $w$ positive when work is done on the system.

Question 2 of 20

An isolated system exchanges with its surroundings:

energy only

matter only

both

neither

Explanation: An isolated system exchanges neither matter nor energy.

Question 3 of 20

Which is NOT a state function?

$U$

$H$

$q$

$S$

Explanation: Heat $q$ is a path function.

Question 4 of 20

Enthalpy is defined as:

$H=U-PV$

$H=U+PV$

$H=U+TS$

$H=U-TS$

Explanation: $H=U+PV$.

Question 5 of 20

For $\Delta n_g=-1$, $\Delta H$ equals:

$\Delta U+RT$

$\Delta U-RT$

$\Delta U$

$\Delta U+2RT$

Explanation: $\Delta H=\Delta U+\Delta n_g RT=\Delta U-RT$.

Question 6 of 20

Mayer’s relation for an ideal gas is:

$C_p-C_v=R$

$C_p+C_v=R$

$C_v-C_p=R$

$C_pC_v=R$

Explanation: $C_p-C_v=R$.

Question 7 of 20

Work done at constant volume is:

$P\Delta V$

$-P\Delta V$

zero

$nRT$

Explanation: At constant volume $\Delta V=0$, so $w=0$.

Question 8 of 20

The standard enthalpy of formation of $\text{O}_2(g)$ is:

$+205$ kJ

$-205$ kJ

zero

$+498$ kJ

Explanation: Elements in their reference state have $\Delta_f H^\circ=0$.

Question 9 of 20

Hess’s law follows from the fact that enthalpy is a:

path function

state function

extensive only quantity

colligative property

Explanation: $\Delta H$ depends only on initial and final states.

Question 10 of 20

Enthalpy of combustion is always:

positive

negative

zero

variable in sign

Explanation: Combustion in oxygen releases heat, so $\Delta_c H<0$.

Question 11 of 20

Enthalpy of neutralisation of a strong acid and strong base is about:

$-57.1$ kJ/mol

$-285.8$ kJ/mol

$+57.1$ kJ/mol

$-110.5$ kJ/mol

Explanation: Only $\text{H}^++\text{OH}^-\rightarrow\text{H}_2\text{O}$ occurs, giving about $-57.1$ kJ/mol.

Question 12 of 20

Lattice enthalpy is obtained using a:

bomb calorimeter

Born–Haber cycle

Carnot cycle

coffee-cup calorimeter

Explanation: It cannot be measured directly; the Born–Haber (Hess) cycle is used.

Question 13 of 20

Using bond enthalpies, $\Delta_r H$ equals:

formed − broken

broken − formed

broken + formed

broken × formed

Explanation: Energy is absorbed breaking and released forming bonds.

Question 14 of 20

Entropy change for a reversible process is:

$q_{rev}T$

$T/q_{rev}$

$q_{rev}/T$

$q_{rev}+T$

Explanation: $\Delta S=q_{rev}/T$.

Question 15 of 20

A spontaneous process has:

$\Delta S_{total}>0$

$\Delta S_{total}<0$

$\Delta S_{total}=0$

$\Delta S_{sys}>0$ always

Explanation: The second law: total entropy of the universe increases.

Question 16 of 20

Gibbs energy change at constant $T,P$ is:

$\Delta H+T\Delta S$

$\Delta H-T\Delta S$

$\Delta U-T\Delta S$

$T\Delta S-\Delta H$

Explanation: $\Delta G=\Delta H-T\Delta S$.

Question 17 of 20

A reaction is spontaneous when:

$\Delta G>0$

$\Delta G=0$

$\Delta G<0$

$\Delta H>0$

Explanation: Spontaneity at constant $T,P$ requires $\Delta G<0$.

Question 18 of 20

The relation $\Delta G^\circ=-RT\ln K$ shows that if $\Delta G^\circ<0$ then:

$K<1$

$K=1$

$K>1$

$K=0$

Explanation: Negative $\Delta G^\circ$ gives $\ln K>0$, so $K>1$ (products favoured).

Question 19 of 20

Entropy of a perfect crystalline solid at 0 K is:

zero

maximum

negative

equal to $R$

Explanation: Third law of thermodynamics.

Question 20 of 20

For $\Delta H<0$ and $\Delta S<0$, the reaction is spontaneous:

at all temperatures

only at low temperature

only at high temperature

never

Explanation: The $-T\Delta S$ term is positive; $\Delta G<0$ only when $T$ is small.

Online Test — Thermodynamics