{"id":515,"date":"2017-11-23T12:25:27","date_gmt":"2017-11-23T12:25:27","guid":{"rendered":"http:\/\/chemdictionary.org\/?p=515"},"modified":"2017-11-23T12:32:55","modified_gmt":"2017-11-23T12:32:55","slug":"hesss-law","status":"publish","type":"post","link":"https:\/\/chemdictionary.org\/hesss-law\/","title":{"rendered":"Hess’s Law"},"content":{"rendered":"

Statement of Hess’s Law<\/h2>\n

Hess’s law of constant heat summation states that the total enthalpy change in a particular reaction is constant regardless whether it occurs in one step or more.<\/p>\n

Explanation of Hess’s Law<\/h2>\n

According to Hess’s law, if A reacts to form the product B, it doesn’t matter how many steps involved to get the product, the total enthalpy change will be same. So the reaction can happen in one step where the enthalpy change is\u00a0\u0394H1<\/sub>. If it occurs in two steps, the enthalpy changes of these two steps are \u0394H2\u00a0<\/sub>and \u0394H3<\/sub>. If it is done in three steps, the enthalpy changes are\u00a0\u0394H4\u00a0<\/sub>, \u0394H5\u00a0<\/sub>and \u0394H6<\/sub>.<\/p>\n

\"See<\/p>\n

So, according to the law,<\/p>\n

\u0394H1\u00a0<\/sub>= \u0394H2\u00a0<\/sub>+ \u0394H3\u00a0<\/sub>= \u0394H4\u00a0<\/sub>+ \u0394H5\u00a0<\/sub>+ \u0394H6\u00a0<\/sub><\/p>\n

Example<\/h2>\n

Formation of carbon dioxide from carbon and oxygen can occur in one step or two steps. In two steps it can be formed via the production of carbon monoxide.<\/p>\n

In one step,<\/p>\n

C(s)<\/sub> + O2(g)<\/sub>\u00a0\u2192 CO2<\/sub>\u00a0\u0394H1<\/sub> = -393 kJ\/mol ……………..(i)<\/p>\n

In two steps,<\/p>\n

step 1: C(s)<\/sub> + 1\/2O2(g)<\/sub>\u00a0\u2192 CO\u00a0\u0394H2<\/sub> = -111 kJ\/mol<\/p>\n

Step 2: CO(g)<\/sub> + 1\/2O2(g)<\/sub>\u00a0\u2192 CO2<\/sub>\u00a0\u0394H3<\/sub> = -282 kJ\/mol
\n_____________________________________
\nOverall change:\u00a0C(s)<\/sub> + O2(g)<\/sub>\u00a0\u2192 CO2<\/sub>\u00a0\u0394H2+3<\/sub> = -393 kJ\/mol =\u00a0\u0394H1<\/sub><\/p>\n

This is same as equation (i) and the heat change also matches with that. So, \u0394H1\u00a0<\/sub>= \u0394H2+3<\/sub>.<\/p>\n

\"By<\/p>\n

Thus it is proved that the Hess’s law is true.<\/p>\n

Applications<\/h2>\n

Using the Hess’s law the enthalpy of a particular reaction can be calculated. Sometime it is impossible to measure in any other way.<\/p>\n

Calculation of enthalpy of formation<\/h3>\n

The enthalpy change of formation can be calculated if we know the entalpy of formation of the reactants and products. As for example, the ethene gas and hydrogen chloride gas reacts to form chloroethane gas.<\/p>\n

C2<\/sub>H4<\/sub>(g)<\/sub> + HCl(g)<\/sub>\u00a0\u2192 C2<\/sub>H5<\/sub>Cl(g)<\/sub>\u00a0\u0394Hf<\/sub> = ? kJ\/mol ……………..(ii)<\/p>\n

The enthalpy change of formation of this reaction need to be calculated. We can calculate that by knowing the heat of formation of ethene gas, hydrogen chloride gas and chloroethane gas from its elements.<\/p>\n

2C(s)<\/sub> + 2H2(g)<\/sub>\u00a0\u2192 C2<\/sub>H4<\/sub>(g)<\/sub>\u00a0\u0394Hf1<\/sub> = +52.2 kJ\/mol……………….(iii)<\/p>\n

1\/2H2(g)<\/sub> + 1\/2Cl2(g)<\/sub>\u00a0\u2192 HCl(g)<\/sub> \u0394Hf2<\/sub>\u00a0= -92.3 kJ\/mol…………(iv)<\/p>\n

2C(s)<\/sub> + 2.5H2(g)\u00a0<\/sub>+ 1\/2Cl2<\/sub> \u2192 C2<\/sub>H5<\/sub>Cl(g)<\/sub>\u00a0<\/sub>\u0394Hf3<\/sub> = -109 kJ\/mol………..(v)<\/p>\n

So, adding (iii) and (iv), we get<\/p>\n

2C(s)<\/sub> + 2.5H2(g)<\/sub>\u00a0+ 2Cl2(g)\u00a0<\/sub>\u2192 C2<\/sub>H4<\/sub>(g)<\/sub>\u00a0<\/sub>+ HCl(g)<\/sub>\u00a0\u0394Hf1+f2<\/sub>\u00a0= -40.1 kJ\/mol…………(vi)<\/p>\n

Then subtracting (vi) from (v), we get<\/p>\n

C2<\/sub>H4<\/sub>(g)<\/sub> + HCl(g)<\/sub>\u00a0\u2192 C2<\/sub>H5<\/sub>Cl(g)<\/sub>\u00a0\u0394Hf<\/sub> = -68.9 kJ\/mol ……………..(vii)<\/p>\n

The equation (ii) and (vii) are same, so the enthalpy change of formation of chloroethane gas from ethene gas and hydrogen chloride gas is\u00a0-68.9 kJ\/mol.<\/p>\n

Calculation of enthalpy of formation from the enthapy of combustion<\/h3>\n

The enthalpy change of formation can be calculated if we know the entalpy of combustion of the reactants and products. As for example, the solid carbon and hydrogen gas reacts to form methane gas.<\/p>\n

C(s)<\/sub> + 2H2(g)<\/sub>\u00a0\u2192 CH4(g)<\/sub>\u00a0\u0394H1<\/sub> = ? kJ\/mol ……………..(viii)<\/p>\n

The enthalpy change of formation of this reaction need to be calculated. We can calculated that by knowing the heat of combustion of carbon, hydrogen and methane.<\/p>\n

C(s)<\/sub> + O2(g)<\/sub>\u00a0\u2192 CO2(g)<\/sub>\u00a0\u0394Hc1<\/sub> = -393.7 kJ\/mol……………….(ix)<\/p>\n

H2(g)<\/sub> + 1\/2O2(g)<\/sub>\u00a0\u2192 H2<\/sub>O(l)<\/sub> \u0394Hc2<\/sub>\u00a0= -285.7 kJ\/mol…………(x)<\/p>\n

CH4(g)<\/sub> + 2O2(g)<\/sub>\u00a0\u2192 CO2(g)<\/sub>\u00a0+ 2H2<\/sub>O(l)\u00a0<\/sub>\u0394Hc3<\/sub> = -890.3 kJ\/mol………..(xi)<\/p>\n

So, doing (ix) + [2 X (x)], we get<\/p>\n

C(s)<\/sub> + 2H2(g)<\/sub>\u00a0+ 2O2(g)\u00a0<\/sub>\u2192 CO2(g)\u00a0<\/sub>+ 2H2<\/sub>O(l)<\/sub> \u0394H2<\/sub>\u00a0= -965.1 kJ\/mol…………(xii)<\/p>\n

Then subtracting (xi) from (xii), we get<\/p>\n

C(s)<\/sub> + 2H2(g)\u00a0<\/sub>\u2192\u00a0CH4(g)\u00a0<\/sub>\u0394H1<\/sub>\u00a0= -74.8 kJ\/mol…………(xiii)<\/p>\n

The equation (viii) and (xiii) are same, so the enthalpy change of formation of methane gas from carbon and hydrogen is\u00a0-74.8 kJ\/mol.<\/p>\n

 <\/p>\n","protected":false},"excerpt":{"rendered":"

Statement of Hess’s Law Hess’s law of constant heat summation states that the total enthalpy change in a particular reaction is constant regardless whether it occurs in one step or more. Explanation of Hess’s Law According to Hess’s law, if A reacts to form the product B, it doesn’t matter how many steps involved to … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"rating_form_position":"","rating_results_position":"","mr_structured_data_type":""},"categories":[22],"tags":[15],"multi-rating":{"mr_rating_results":[{"adjusted_star_result":4.58,"star_result":4.58,"total_max_option_value":5,"adjusted_score_result":4.58,"score_result":4.58,"percentage_result":91.58,"adjusted_percentage_result":91.58,"count":19,"post_id":515}]},"_links":{"self":[{"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/posts\/515"}],"collection":[{"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/comments?post=515"}],"version-history":[{"count":0,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/posts\/515\/revisions"}],"wp:attachment":[{"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/media?parent=515"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/categories?post=515"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/chemdictionary.org\/wp-json\/wp\/v2\/tags?post=515"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}