The entropy change can be calculated by using the expression $∆\mathrm{S}=\frac{{\mathrm{q}}_{\mathrm{rev}}}{\mathrm{T}}$. When water freezes in a glass beaker, the correct statement among the following is:

1.	∆ S  (system) decreases but  ∆ S  (surroundings) remains the same
2.	∆ S  (system) increases but  ∆ S  (surroundings) decreases
3.	∆ S  (system) decreases but  ∆ S  (surroundings) increases
4.	∆ S  (system) decreases but  ∆ S  (surroundings) also decreases

Consider the reactions given below. On the basis of these reactions find out which of the

the algebraic relationships is correct?

$\left(i\right)$ $C\left(g\right)$ $+4H$ $\left(g\right)$ $\to C{H}_4\left(g\right);$ $$
$∆H_r$ $=x$ $kJ$ $mo{l}^{-1}$
$\left(ii\right)$ $C\left(graphite\right)$ $+$ $2H_2$ $\left(g\right)$ $\to$ $$ $C{H}_4;$ $$
$∆H_r$ $$ $=$ $y$ $kJ$ $mo{l}^{-1}$

The enthalpies of elements in their standard states are taken as zero. The enthalpy of formation of a compound is-

The enthalpy of sublimation of a substance is equal to-

1. Enthalpy of fusion + Enthalpy of vaporisation

2. Enthalpy of fusion

3. Enthalpy of vaporisation

4.  Twice the enthalpy of vaporisation

Which of the following is not correct?

1. $∆\mathrm{G}$ is zero for a reversible reaction.

2. $∆\mathrm{G}$ is positive for a spontaneous reaction.

3. $∆\mathrm{G}$ is negative for a spontaneous reaction.

4. $∆\mathrm{G}$ is positive for a non-spontaneous reaction.

A necessary condition for an adiabatic change is-

1. ∆T = 0

2. ∆P = 0

3. q = 0

4. w = 0

The enthalpy of formation of all elements in their standard state is-

$∆\mathrm{U}°$ $$for combustion of methane is –x kJ mol^–1.

The value of $∆\mathrm{H}°$ for the same reaction would be:

$1.$ $=$ $∆\mathrm{U}°$
$2.$ $>$ $∆\mathrm{U}°$
$3.$ $<$ $∆\mathrm{U}°$
$4.$ $=0$

701 J of heat is absorbed by a system and 394 J of work is done by the system. The change in internal energy for the process is-

The reaction of cyanamide, ${\mathrm{NH}}_2\mathrm{CN}$ $\left(\mathrm{s}\right)$ with dioxygen, was carried out in a bomb calorimeter, and ∆U was found to be $-742.7$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$ at 298 K.
\(\small{\mathrm{NH}_2 \mathrm{CN}(\mathrm{s})+\frac{3}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{g})+\mathrm{CO}_2(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{l})}\)

${}_{}$The enthalpy change for the reaction at 298 K would be -

$1.$ $-741.3$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$2.$ $+$ $753.9$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$3.$ $+$ $772.$ $7$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$
$4.$ $-845.$ $1$ $\mathrm{kJ}$ ${\mathrm{mol}}^{-1}$