Hess's Law is crucial in determining enthalpy changes. What fundamental principle does Hess's Law rely on?
IB Diploma Programme (DP) - SL & HL · Chemistry
反應性1.2—反應中的能量循環:练习题
5 道选择题即时批改,另有 5 道文字题附完整解题步骤,全部围绕「反應性1.2—反應中的能量循環」。
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.
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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, \( \text{C}_2\text{H}_5\text{OH}(l) \).
先自己写一遍答案,再对照解题步骤。
State Hess's Law and explain how it allows the determination of enthalpy changes that cannot be measured directly through experimentation.
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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.
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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.
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