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Thermodynamics — Chemistry STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceChemistryThermodynamics5 Marks
Question
  1. Heat(q) and work done (W) individually are not state functions but their sum is always a state function. Explain why?
  2. Calculate the standard enthalpy change $(\Delta_\text{r}\text{H}^\theta)$ and standard internal energy change $(\Delta_\text{r}\text{U}^\theta)$ for the following reaction at 300K:

OF2(g) + H2O(g) + O2(g) + 2HF(g)

Standard enthalpy of formation $(\Delta_\text{r}\text{H}^\theta)$ of e n various species are given below:

$\Delta_\text{f}\text{H}^\theta\text{kJ mol}^{-1}:$ OF2(g) = 23.0, H2O(g)= -241.8, HF(g) = -268.6, R = 8.314 JK-1mol-1.

✓

Answer

  1. 'q’ is not a state function because it depends upon path.

'w' is not a state function because it depends upon path.

$\text{q}+\text{w}=\Delta\text{U}$ which is a state function because it is independent of path.

  1. $\text{OF}_2(\text{g})+\text{H}_2\text{O(g)}\overrightarrow{\ \ \ \ \ \ \ }\ \text{O}_2(\text{g})+2\text{HF(g)}$

$\Delta\text{n}=1+2-2=1$

$\Delta_\text{r}\text{H}^\circ=\sum\Delta_\text{f}\text{H}^\circ(\text{products})-\sum\Delta_\text{f}\text{H}^\circ(\text{reactants})$

$=\Delta_\text{f}\text{H}^\circ(\text{O}_2)+2\Delta_\text{f}\text{H}^\circ(\text{HF})-\Delta_\text{f}\text{H}^\circ(\text{OF}_2)\\-\Delta_\text{f}\text{H}^\circ(\text{H}_2\text{O})$

$=0+2\times(-268.6)-23.0-(-241.8)$

$=537.2-23.0+241.8$

$=-318.4\text{kJ}$

$\Delta_\text{f}\text{H}^\circ=\Delta_\text{r}\text{U}^\circ+\Delta\text{nRT}$

$-318.4\text{kJ}=\Delta_\text{r}\text{U}^\circ+\frac{1\times8.314\times300}{1000}\text{kJ}$

$-318.4\text{kJ}=\Delta_\text{r}\text{U}^\circ+2.4942\text{kJ}$

$\Delta_\text{r}\text{U}^\circ=-320.8942\text{kJ}$

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