The following acid-base reaction is performed in a coffee cup calorimeter:
H+(aq) + OH-(aq) → H2O(l)
The temperature of 110 g of water rises from 25.0°C to 26.2°C when 0.10 mol of H+ is reacted with 0.10 mol of OH-
Calculate q water (heat of water)
Calculate ΔH if 1.00 mol OH- reacts with 1.00 mol H+
*After calculating, q water is 550J
the solution given for the next question is:
We know that when 0.010 mol of H+ or OH- reacts, ΔH is - 550 J:
0.010 mol H+ ~ -550 J
Therefore, for 1.00 mol of H+ (or OH-):
ΔH = 1.00 mol H+ x (-550 J / 0.010 mol H+)
ΔH = -5.5 x 104 J
ΔH = -55 kJ
*My question is, why in the solution it is said that 0.01mol of H+ is reacted, and not 0.1mol, as in the question? Is it a typo error or is it supposed to be that way?
*And also, the q water is the same as the q reaction right?
H+(aq) + OH-(aq) → H2O(l)
The temperature of 110 g of water rises from 25.0°C to 26.2°C when 0.10 mol of H+ is reacted with 0.10 mol of OH-
Calculate q water (heat of water)
Calculate ΔH if 1.00 mol OH- reacts with 1.00 mol H+
*After calculating, q water is 550J
the solution given for the next question is:
We know that when 0.010 mol of H+ or OH- reacts, ΔH is - 550 J:
0.010 mol H+ ~ -550 J
Therefore, for 1.00 mol of H+ (or OH-):
ΔH = 1.00 mol H+ x (-550 J / 0.010 mol H+)
ΔH = -5.5 x 104 J
ΔH = -55 kJ
*My question is, why in the solution it is said that 0.01mol of H+ is reacted, and not 0.1mol, as in the question? Is it a typo error or is it supposed to be that way?
*And also, the q water is the same as the q reaction right?