Respuesta :
25.0 mL of 0.500 M H₂SO₄ is added to 25.0 mL of 1.00  M KOH in a coffee-cup calorimeter at 23.50°C, the temperature rises to 30.17°C.
DH of this reaction is 55.8 kJ/m H2O .
Solution ;
{25.0 mL x. ( 1L / 1000mL )} × {( 0.500 mol H2SO4 )/1L }
= 0.0125 mol H2SO4
{25.0 mL x. ( 1L / 1000mL )} × {( 1.00 mol KOH )/1L }
= 0.0250 mol KOH
Equation: Hâ‚‚SO, +2KOH Kâ‚‚SO, + 2Hâ‚‚O
The two reactants are present in stoichiometric proportions and are present in equal amounts in the reaction. So either of the original moles will yield the following moles of water:
0.0125 mol H2SO4 × ( 2 mol H₂O/1m H2SO4 ) = 0.0250 mol H2O
0.0125 mol NaOH × ( 2 mol H₂O/1m KOH ) = 0.0250 mol H2O
So, 0.0250 moles H20 has produced.
Volume of water = (25.0+25.0) mL = 50.0 mL
The heat absorbed by calorimeter contents (water)is:
q = { 50.0mL × ( 1.00g / ml ) } × { 4.184J/ g° C × ( 30.17 - 23.50 ) ° } = 1395 J = 1.396KJ
Volume of water = (25.0+25.0) mL = 50.0 mL
The heat absorbed by calorimeter contents (water) is :
q cal = { 50.0 mL x (1.00 g /mL) } ( 4.184J /g° C ) × (30.17-23.50)°C } = 1395 J = 1.395 KJ
Heat of reaction, q calories = (-) 1.395 KJ
Heat of neutralization per mole of water, ∆H = { ( - ) 1.395 kJ / 0.0250 mol H₂O } = 55.8 kJ/m H2O
What  is Specific Heat ?
The amount of heat needed to raise a substance's temperature by one degree Celsius in one gram, also known as specific heat. Typically, calories or joules per gram per degree Celsius are used as the units of specific heat. For instance, water has a specific heat of 1 calorie (or 4.186 joules) per gram per degree Celsius.
To know more about Specific heat please click here ; https://brainly.com/question/21406849
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