Reactivity 1.2—Energy cycles in reactions: แบบฝึกหัด

Hess's Law is crucial in determining enthalpy changes. What fundamental principle does Hess's Law rely on?

Given the following data:
\( ΔH_{\text{at}}^\theta [Na(s)] = +107 \text{ kJ mol}^{-1} \)
\( IE_1 [Na(g)] = +496 \text{ kJ mol}^{-1} \)
\( ΔH_{\text{at}}^\theta [\frac{1}{2}Cl_2(g)] = +122 \text{ kJ mol}^{-1} \)
\( EA_1 [Cl(g)] = -349 \text{ kJ mol}^{-1} \)
\( ΔH_f^\theta [NaCl(s)] = -411 \text{ kJ mol}^{-1} \)
What is the lattice enthalpy of formation of NaCl(s) in \( \text{kJ mol}^{-1} \)?

Lithium chloride (\(\text{LiCl}\)) has a more exothermic enthalpy of solution than sodium chloride (\(\text{NaCl}\)). Which statement best explains this observation using an energy cycle?

Which enthalpy change is represented by the equation below?
\( K^+(g) + Cl^-(g) \rightarrow KCl(s) \)

The standard enthalpy changes of combustion for carbon, hydrogen, and ethene are given below:
\( C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_c^\theta = -394 \text{ kJ mol}^{-1} \)
\( H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H_c^\theta = -286 \text{ kJ mol}^{-1} \)
\( C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l) \quad \Delta H_c^\theta = -1411 \text{ kJ mol}^{-1} \)
What is the standard enthalpy of formation of ethene, \( C_2H_4(g) \), in \( \text{kJ mol}^{-1} \)?

Given that \(A \rightarrow B\) has \(\Delta H_1 = +40\text{ kJ mol}^{-1}\) and \(C \rightarrow B\) has \(\Delta H_2 = +65\text{ kJ mol}^{-1}\), determine the enthalpy change for the reaction \(A \rightarrow C\) using Hess's Law.

Using the standard enthalpy of combustion values for \( \text{C}(s) \) (\( -394\text{ kJ mol}^{-1} \)), \( \text{H}_2(g) \) (\( -286\text{ kJ mol}^{-1} \)), and \( \text{C}_2\text{H}_5\text{OH}(l) \) (\( -1367\text{ kJ mol}^{-1} \)), calculate the standard enthalpy of formation for ethanol, \( \text{C}_2\text{H}_5\text{OH}(l) \).

State Hess's Law and explain how it allows the determination of enthalpy changes that cannot be measured directly through experimentation.

Using Hess's Law and the standard enthalpy of combustion data provided below, calculate the enthalpy change for the conversion of graphite to diamond:
\(C(\text{s, graphite}) \rightarrow C(\text{s, diamond})\)

Data:
1. \(C(\text{s, graphite}) + O_2(g) \rightarrow CO_2(g)\)    \(\Delta H_c^{\circ} = -393.5\text{ kJ mol}^{-1}\)
2. \(C(\text{s, diamond}) + O_2(g) \rightarrow CO_2(g)\)    \(\Delta H_c^{\circ} = -395.4\text{ kJ mol}^{-1}\)

Construct an energy cycle to show your reasoning and state whether the process is exothermic or endothermic.

The enthalpy of solution for calcium chloride can be calculated using hydration and lattice enthalpies. Given:
Lattice enthalpy (\(\Delta H_{lat}^{\ominus}\)) of \(CaCl_2\): \(-2258\text{ kJ mol}^{-1}\)
Hydration enthalpy (\(\Delta H_{hyd}^{\ominus}\)) of \(Ca^{2+}(g)\): \(-1577\text{ kJ mol}^{-1}\)
Hydration enthalpy (\(\Delta H_{hyd}^{\ominus}\)) of \(Cl^{-}(g)\): \(-363\text{ kJ mol}^{-1}\)
(a) State the equation for the lattice enthalpy of \(CaCl_2\) as defined by the IB.
(b) Construct an energy cycle and calculate the enthalpy of solution for \(CaCl_2(s)\).
(c) Suggest why the entropy change of the system is positive when \(CaCl_2\) dissolves in water.