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Example 5.1. Calculating heat of a reactionProblem:The thermochemical equation for the reaction of zinc with sulfur is
Calculate the heat, liberated when 8.0 g of sulfur react according to this equation. Solution: The DHr given is the energy liberated when 1 mol S reacts. The molar mass of S is 32.0 g/mol, so 32.0 g S ® – 250.4 kJ 8.0 g S ® x kJ x = 8.0×(– 250.4)/32.0 = – 63 kJ Answer: in reaction of 8.0 g S with zinc 63 kJ are liberated. Standard conditions. Most thermochemical equations are written for the standard conditions (298 K and 1 atm) applied both to reagents and products. Heat effect of a reaction under the standard conditions is denoted as DHr°. If the temperature and pressure are different from 298 K and 1 atm they should be specified in the thermochemical equation. Standard state of a substance is its only or the most stable state at standard conditions. Under these conditions a substance in its most stable aggregate state (liquid for H2O) and most stable allotropic form (O2 for oxygen) or crystalline polymorph (graphite for C) is said to be in a standard state. The standard enthalpy of formation of a substance, DHf°, is defined as the heat associated with the formation of one mole of the compound from elementary substancesin their standard states. For example, the standard enthalpy of liquid water formation is the heat of reaction: H2(g) + ½ O2(g) → H2O(l); DH °reaction = DHf ° H2O(l) = –286 kJ In accordance with the definition of the DHf°,enthalpies of the elementary substances in their standard statesare equal to zero. The standard enthalpy of formation is a fundamental property of any compound. Most chemistry textbooks contain tables of DHf ° values. Calculating of heat effects of reactions. Laws of thermochemistry are based on the fundamental law of conservation of matter[4]. Heat effects of reactions including supposed or even impossible ones can be calculated using the laws of thermochemistry. The first law of thermochemistrystates that heat effect of a reaction is equal in magnitude but opposite in sign to ΔH for the reverse reaction. For example:
The second law of thermochemistryalso calledHess's lawstates that ΔHr is the same whether a reaction occurs in one step or in a series of steps. For example, excess of phosphorous react with chlorine to form PCl3. Then PCl3 can react with excess of chlorine to form PCl5. The phosphorous pentachloride can be also synthesized in one step
As ΔHr must be independent of the path of a reaction, ΔH3 = ΔH1 + ΔH2 = ΔH1 + ΔH2 = – 280 + (– 87) = – 367 kJ Hess's lawcan be applied to calculate the enthalpy change that can not be measured directly. For example, both graphite and diamond (the two crystalline forms of carbon) can be oxidized by an excess of oxygen to carbon dioxide. C(graphite) + O2(g) → CO2(g) ΔH° = –393.51 kJ C(diamond) + O2(g) → CO2 ΔH° = –395.40 kJ The standard enthalpy changes for these reactions can be combined to obtain ΔH° for the phase transformation between graphite and diamond. Subtraction of the second reaction from the first (i.e., writing the second equation in reverse and adding it to the first one) yields C(graphite) → C(diamond) ΔH° = 1.89 kJ Hess’ law allows predicting the change in the enthalpy of huge numbers of chemical reactions from a relatively small base of experimental data. Date: 2015-12-11; view: 865
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PCl3
