Acommon laboratory reaction is the neutralization of an acid with a base. when 50.0 ml of 0.500 m hcl at 25.0°c is added to 50.0 ml of 0.500 m naoh at 25.0°c in a coffee cup calorimeter, the temperature of the mixture rises to 28.2°c. assume the mixture has a specific heat capacity of 4.18 j/(g. k) and the densities of all solutions are 1.00 g/ml

a.) calculate the qrxn
b.) calculate deltah rxn per mole of acid
c.) when this reaction is repeated with 50.0 ml of 2.0 m hci at 25.0°c, the temperature of the mixture also increases to 28.2°c. explain

a) qrxn = 1,337.6 J

b)  ΔHrxn /mol = -5.35 kJ/mol

c) See explanation below

Explanation:

a)  q rxn = mcΔT where m is the mass of water, c  its specific heat capacity , and ΔT is the change in temperature.

m H₂O = (50.0 mL + 50.0 mL) 1.00 g/mL= 100 g

q rxn = 100 g x 4.18 J/gK x (28.2 -25.0) K  = 1,337.6 J

b) ΔHrxn = - qrxn

mol HCl =( 50mL/1000mL/L )x  0.500 mol/L = 0.025 mol

ΔHrxn / mol = -1337.6 J/0.025 mol = -53,504.0 J = -5.35 kJ/mol

c)   We just calculated the heat of neutralization per mol HCl, but thats not the whole story.

We have to ensure that if we increase the quantity of HCl ( 0.100 mol) we also increase the amount NaOH in the same proportion, i.e. 0.100, if not only 0.025 mol HCl will react liberating the same amount of heat calculated above.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml

може ли да зареждате повече информация, моля