Question

11£®º¬ÓÐ11.2g KOHµÄÏ¡ÈÜÒºÓë1L 0.1mol/LµÄH₂SO₄ÈÜÒº·´Ó¦£¬·Å³ö11.46kJµÄÈÈÁ¿£¬±íʾ¸Ã·´Ó¦ÖкÍÈȵĻ¯Ñ§·½³ÌʽΪ£¨¡¡¡¡£©

• A£® KOH£¨aq£©+$\frac{1}{2}$H₂SO₄£¨aq£©¨T$\frac{1}{2}$K₂SO₄£¨aq£©+H₂O£¨l£©¡÷H=-11.46 kJ/mol
• B£® 2KOH£¨aq£©+H₂SO₄£¨aq£©¨TK₂SO₄£¨aq£©+2H₂O£¨l£©¡÷H=-11.46 kJ/mol
• C£® 2KOH£¨aq£©+H₂SO₄£¨aq£©¨TK₂SO₄£¨aq£©+2H₂O£¨l£©¡÷H=-114.6 kJ/mol
• D£® KOH£¨aq£©+$\frac{1}{2}$H₂SO₄£¨aq£©¨T$\frac{1}{2}$K₂SO₄£¨aq£©+H₂O£¨l£©¡÷H=-57.3 kJ/mol

Answer 1

·ÖÎö n£¨KOH£©=$\frac{11.2g}{56g/mol}$=0.2mol£¬n£¨H₂SO₄£©=1L¡Á0.1mol/L=0.1mol£¬¶þÕß·¢Éú2KOH£¨aq£©+H₂SO₄£¨aq£©¨TK₂SO₄£¨aq£©+2H₂O£¬Éú³É0.2molË®£¬ÔòÉú³É1molË®·Å³ö57.3 kJÈÈÁ¿£¬ÒÔ´ËÊéд»¯Ñ§·½³Ìʽ£®

½â´ð ½â£ºn£¨KOH£©=$\frac{11.2g}{56g/mol}$=0.2mol£¬n£¨H₂SO₄£©=1L¡Á0.1mol/L=0.1mol£¬¶þÕß·¢Éú2KOH£¨aq£©+H₂SO₄£¨aq£©¨TK₂SO₄£¨aq£©+2H₂O£¬Éú³É0.2molË®£¬ÔòÉú³É1molË®·Å³ö57.3 kJÈÈÁ¿£¬
ÖкÍÈÈΪÉú³É1molË®·Å³öµÄÈÈÁ¿£¬Ôò¸Ã·´Ó¦ÖкÍÈȵĻ¯Ñ§·½³ÌʽΪKOH£¨aq£©+$\frac{1}{2}$H₂SO₄£¨aq£©¨T$\frac{1}{2}$K₂SO₄£¨aq£©+H₂O£¨l£©¡÷H=-57.3 kJ/mol£®
¹ÊÑ¡D£®

µãÆÀ ±¾Ì⿼²éÈÈ»¯Ñ§·½³ÌʽµÄÊéд·½·¨ºÍ×¢ÒâÎÊÌ⣬Ϊ¸ßƵ¿¼µã£¬²àÖØ¿¼²éѧÉúµÄ·ÖÎö¡¢¼ÆËãÄÜÁ¦£¬¹Ø¼üÊǶÔÓ¦»¯Ñ§·½³Ìʽ¼ÆÁ¿ÊýÌõ¼þϵÄìʱä¼ÆË㣬ÌâÄ¿½Ï¼òµ¥£®