InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
a student investigated temperature changes during a reaction between citric acid and sodium hydrogencarbonate solution. HOW do the results in table 4 show that the reaction is endothermic? |
| Answer» <html><body><p><strong>Answer:</strong></p><p><<why acid="" and="" between="" carbonate="" endothermic="" hydrochloric="" hydrogen="" is="" reaction="" sodium="" the="">></why></p><p></p><p>The question is undefined!</p><p></p><p>You don't specifiy if NaHCO3 is <a href="https://interviewquestions.tuteehub.com/tag/solid-1216587" style="font-weight:bold;" target="_blank" title="Click to know more about SOLID">SOLID</a> or acqueous (dissolved in water) but, mostly, you don't specifiy if hydrochloric acid is gaseous or in water solution.</p><p></p><p>Furthermore, if the reaction happens in a container with excess water, the NaCl formed could be considered solid or acqueous (it depends on how much water there is and how much you wait...)</p><p></p><p>So:</p><p></p><p>ΔHr=−411−286−394−(−951−166)= </p><p></p><p>=<a href="https://interviewquestions.tuteehub.com/tag/26-298265" style="font-weight:bold;" target="_blank" title="Click to know more about 26">26</a> kJ mol−1 </p><p></p><p>It's a positive value so it's endothermic.</p><p></p><p>The reason for this is that in order for the reaction to happen, you have to give energy to break:</p><p></p><p>The ionic lattice Na++HCO−3 </p><p>One C−O <a href="https://interviewquestions.tuteehub.com/tag/bond-900583" style="font-weight:bold;" target="_blank" title="Click to know more about BOND">BOND</a> in HCO−3 </p><p>One H−Cl bond in <a href="https://interviewquestions.tuteehub.com/tag/hcl-479502" style="font-weight:bold;" target="_blank" title="Click to know more about HCL">HCL</a> </p><p>but, on the other end, you get energy only by forming of:</p><p></p><p>The ionic lattice Na++Cl− </p><p>One H−O bond in H2O (the other H−O bond was already present in HCO−3 )</p><p>so the net result is a loss of energy.</p><p></p><p>Now let's see what would be with gaseous HCl ; since its ΔH of formation is −92 kJ mol−1: </p><p></p><p>ΔHr=−411−286−394−(−951−92)= </p><p></p><p>=−48 kJ mol−1 </p><p></p><p>Since it's negative, it's exothermic could be exothermic instead of endothermic!</p><p></p><p>I imagine you refer to solid NaHCO3, NaCl and acqueous HCl, so as follows.</p><p></p><p></p></body></html> | |