Question

Three mole of Ag is heated from 300 K to 1000 K. Calculate ΔH when P = 1 atm and ${\mathrm{C}}_{\mathrm{P}}=23+0.01\mathrm{T}.$

• A. 62 kJ/mol
• B. 45 kJ/mol
• C. 38 kJ/mol
• D. 54 kJ/mol

Answer 1

The correct option is A

62 kJ/mol

Explanation for correct answer:-

Option (A) 62 kJ/mol

Step 1: Given data

Moles of Ag= 3

${\mathrm{C}}_{\mathrm{P}}=23+0.01\mathrm{T}.$

Pressure P= 1 atm

Initial temperature ${\mathrm{T}}_1$= 300K

Final temperature ${\mathrm{T}}_2$= 1000K

Step 2: Calculating the enthalpy

We know that,

$$\begin{gathered} ∆\mathrm{H}={\int }_{{\mathrm{T}}_1}^{{\mathrm{T}}_2}{\mathrm{nC}}_{\mathrm{p}}d\mathrm{T} \\ \Rightarrow ∆\mathrm{H}=\mathrm{n}{\int }_{300}^{1000}(23+0.01)d\mathrm{T} \\ \Rightarrow ∆\mathrm{H}=3{\left[23\mathrm{T}+\frac{0.01{\mathrm{T}}^2}{2}\right]}_{300}^{1000} \\ \Rightarrow ∆\mathrm{H}=3\left[2\times 23\times (1000-300)+\frac{0.01\left\{{\left(1000\right)}^2-{\left(300\right)}^2\right\}}{2}\right] \\ \Rightarrow ∆\mathrm{H}=61950\mathrm{J}{\mathrm{mol}}^{-1} \\ \Rightarrow ∆\mathrm{H}\approx 62\mathrm{KJ}{\mathrm{mol}}^{-1} \end{gathered}$$

Therefore, option (A) 62 kJ/mol is the correct answer