50ml of 0.10M HCL is mixed with 50ml of 0.10M NAOH.The solution of the temperature rises by 276K.Calculate the enthalpy of neutralization per mole of HCL.
 
1)-2.5*100 KJ
2)-1.3*100KJ
3)-8.4*10KJ
4)-6.3*10KJ

Asked by TVISHA Bhatt | 17th Aug, 2014, 03:32: PM

Expert Answer:

The equation for the reaction is
 
Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

1 mole of acid reacts with 1 mole of alkali to form 1 mole of water.

Number of moles of acid used = 0.05 x 0.1 = 0.005

Number of moles of alkali used = 0.05 x 0.1 = 0.005

So number of moles of water formed = 0.005
 
Use formula,
 
Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.
c = 4.18 kJ / kg °C, m = mass of mixed solution = 50 + 50 = 100 /1000 = 0.1 kg,Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. = 3 °C
 
so, Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. = - 4.18 x 0.1 x 3 = 1.254 kJ
 
0.005 moles Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. 1.254 kJ
 
1 mole Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. = - 2.5 x 102 kJ                                                                   

Answered by Arvind Diwale | 19th Aug, 2014, 11:24: AM