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

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