A heat capacity of water = heat capacity of calorimeter, the heat gained by water = heat lost by calorimeter
\(=125\times (296.5-286.4)\times 4.184 \ J=5282J=5.282kJ\)

Use the following steps to solve out such problems.
Step I   Write the balanced equation
\({ NH }_{ 2 }CN(s)+\frac { 3 }{ 2 } { O }_{ 2 }(g)\rightarrow { N }_{ 2 }(g)+{ CO }_{ 2 }(g)+{ H }_{ 2 }O(l)\)
Step II  Calculate \({ \Delta n }_{ g }\)
Difference of moles of gaseous products and
reactants, \({ \Delta n }_{ g }={ n }_{ p }-{ n }_{ r }=2-\frac { 3 }{ 2 } =\frac { 1 }{ 2 } =0.5mol\)
Step III Calculate \(\Delta H\) by using the formula
\(\Delta H=\Delta U+{ \Delta n }_{ g }RT\)
\(\Delta H=-742.7 \ kj \ { mol }^{ -1 }\)
\( +(0.5mol\times 8.314\times { 10 }^{ -3 }kj \ { mol }^{ -1 }\times 298K)\)
During calculation always remember the units of different quantities, i.e. 
\(\Delta H,\Delta U\) and R must be the same.

\(For \ isothermal \ reversible \ expansion \ of \ ideal \ gas\)
\({ W }_{ rev }=-2.303nRT \ log\frac { { V }_{ 2 } }{ { V }_{ 1 } }\)
\( =-2.303\times 5.2\times 8.314\times 298 \ log \ \left( \frac { 127.05 }{ 42.35 } \right)\)
\( =14156.38J\)

The standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states (reference states) is called standard molar enthalpy of formation, \({ \Delta }_{ f }{ H }^{ o }\) .
\(Ca(s)+C(s)+\frac { 3 }{ 2 } { O }_{ 2 }(g)\longrightarrow Ca{ CO }_{ 3 }(s);{ \Delta }_{ f }{ H }^{ o }\)
This reaction is different from the given reaction.
Hence, \({ \Delta }_{ r }{ H }^{ o }\neq { \Delta }_{ f }{ H }^{ o }\)

Enthalpy change, internal energy change and pressure

\(w=-2.303nRT \ log\frac { { V }_{ 2 } }{ { V }_{ 1 } } -4754\)
\(=-2.303\times n\times 8.314\times 300\times log\frac { 20 }{ 10 }\)
\(On \ solving, \ n=2.75mol\)

Hess's law of heat summation.

\({ \triangle }_{ f }{ H }^{ \circ }\) for Cl₂(g) will be zero but \({ \triangle }_{ f }{ H }^{ \circ }\) for Br₂ (g) will not be zero because liquid bromine is its elementry state and not gaseous bromine.

The sign of work done will be positive, i.e. work will be done on the system due to a decrease in volume.

For the given reaction,
2A(g)+B(g)→2D(g)
Δng=2−(3)=–1 mole
Substituting the value of ΔUθ the expression of Δ H:
Δ H^θ=Δ U^θ+ ΔngRT
=(−10.5kJ)+(−1)(8.314×10⁻³kJ K⁻¹mol⁻¹)(298 K)
=−10.5 kJ⁻².48 kJ
ΔH^θ=−12.98 kJ
Substituting the value of ΔH^θ and ΔS^θ in the expression of ΔG^θ:
ΔG^θ=ΔH^θ=TΔS^θ
=−12.98 kJ−(298K)(−44.1JK⁻¹)
=−12.98 kJ+13.14 kJΔ G^θ= +0.16kJ
Since Δ Gθ for the reaction is positive, the reaction will not occur spontaneously.

The conversion of vapour into liquid is condensation. The enthalpy of condensation is negative of enthalpy of vaporisation.
 \(\triangle _{ vap }{ H }^{ o }=-\triangle _{ cond }{ H }^{ o }\)
For condensation of diethyl ether (i.e., conversion of vapour to liquid)
\(\triangle _{ cond }{ S }^{ o }=\frac { \triangle _{ cond }{ H }^{ o } }{ T } =\frac { -26.0\times1{ 0 }^{ 3 }{ mol }^{ -1 } }{ 308 } \)
\( =-84.4J{ K }^{ -1 }{ mol }^{ -1 }\)