Question

For silver, \[{\text{Cp}}\left( {{\text{J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}}
\right){\text{ = 23 + 0}}{\text{.01T}}\] . If the temperature (T) of \[{\text{3 moles}}\] of silver is raised from \[300{\text{ K}}\] to \[1000{\text{ K}}\] at \[1{\text{ atm}}\] pressure, the value of \[\Delta H\] will be close to:
A \[21{\text{ kJ}}\]
B \[16{\text{ kJ}}\]
C \[13{\text{ kJ}}\]
D \[62{\text{ kJ}}\]

Answer 1

Hint: We can begin by writing an expression to obtain \[\Delta H\] from the heat capacity
\[\Delta H = n\int\limits_{{T_1}}^{{T_2}} {{C_{p,m}}dT} \]
Then, you should substitute values in the above expression:
In the next step, you integrate the above expression within the limits of the given temperature range, and further solve the expression to obtain \[\Delta H\]value.


Complete Step by step answer: The enthalpy change and the heat capacity are mathematically related to each other.
Write an expression to obtain \[\Delta H\] from the heat capacity
\[\Delta H = n\int\limits_{{T_1}}^{{T_2}} {{C_{p,m}}dT} \]
Substitute the number of moles as 3, the initial and final temperatures as 300 and 1000 and the heat capacity of silver \[{\text{Cp}}\left( {{\text{J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}} \right)\] as \[{\text{ 23 + 0}}{\text{.01T}}\] in the above expression:
\[\Delta H = 3 \times \int\limits_{300}^{1000} {\left( {{\text{23 + 0}}{\text{.01T}}} \right)dT} \]
Integrate the above expression
\[
 \Rightarrow \Delta H = 3 \times \left[ {23\left( {\text{T}} \right)_{{{\text{T}}_1}}^{{{\text{T}}_2}} + \dfrac{{0.01}}{2}\left( {{{\text{T}}^2}} \right)_{{{\text{T}}_1}}^{{{\text{T}}_2}}} \right] \\
  \Rightarrow \Delta H = 3 \times \left[ {23\left( {{{\text{T}}_2}{\text{ }} - {\text{ }}{{\text{T}}_1}} \right) + \dfrac{{0.01}}{2}\left( {{{\text{T}}_2}^2{\text{ }} - {\text{ }}{{\text{T}}_1}^2} \right)} \right] \\
 \Rightarrow \Delta H = 3 \times \left[ {23\left( {{\text{1000 }} - {\text{ 300}}} \right) + \dfrac{{0.01}}{2}\left( {{\text{100}}{{\text{0}}^2}{\text{ }} - {\text{ 30}}{{\text{0}}^2}} \right)} \right] \\
 \]
Further solve the above expression
\[
  \Delta H = 3 \times \left[ {23\left( {{\text{700}}} \right) + \dfrac{{0.01}}{2}\left( {{\text{1,000,000 }} - {\text{ 90,000}}} \right)} \right] \\
   \Rightarrow \Delta H = 3 \times \left[ {16100 + \dfrac{{0.01}}{2}\left( {{\text{910,000 }}} \right)} \right] \\
 \Rightarrow \Delta H = 3 \times \left[ {16100 + \dfrac{{9100}}{2}} \right] \\
 \Rightarrow \Delta H = 3 \times \left[ {16100 + 4550} \right] \\
 \]

Further solve the above expression
\[
  \Delta H = 3 \times 20650 \\
  \Rightarrow \Delta H = 61950{\text{ kJ}} \\
 \Rightarrow \Delta H \simeq 62{\text{ kJ}} \\
 \]
Hence, the value of the enthalpy change \[\Delta H\] for the reaction is \[62{\text{ kJ}}\] .

Hence, the option (D) is the correct option.

Note: Students should keep in mind some procedures such that, solve a definite integral according to the following formula.
\[
  \int\limits_i^f {\left( {{\text{a + bx}}} \right)d{\text{x}}} = \left[ {a\left( {\text{x}} \right)_i^f + b\left( {{{\text{x}}^2}} \right)_i^f} \right] \\
  \int\limits_i^f {\left( {{\text{a + bx}}} \right)d{\text{x}}} = \left[ {a\left( {{{\text{x}}_f} - {{\text{x}}_i}} \right) + b\left( {{{\text{x}}_f}^2 - {{\text{x}}_i}^2} \right)} \right] \\
 \]
Here, you have integrated the function \[\left( {{\text{a + bx}}} \right)\] between the limits i and f . i and f represents initial and final values for the variable x. a and b are constants