Question

A 50.0 ml sample of a 1.00 m solution of cuso4 is mixed with 50.0 ml of 2.00 m koh in a calorimeter. cuso4 (1m) + 2koh (2m) cu(oh)2(s) + k2so4 (0.5m). the temperature of both solutions was 20.2 °c before mixing and 26.3 °c after mixing. the heat capacity of the calorimeter is 12.1 j/°c. assume the specific heat and density of the solution after mixing are the same as those of pure water. from this data, calculate theδh for the process if there is 0.05 mols of cuso4. (energy of the water + energy of the calorimeter)/(1000 x mol)= kj/mol of reaction

Answer 1

Answer:- $\Delta H$ = -$\frac{52.52kJ}{mol}$

Solution:- Total volume of the solution = 5.0 mL + 50.0 mL = 100.0 mL

Density of solution is same as of water that is 1.0 g per mL. So, the mass of solution, m = $100.0mL(\frac{1.0g}{mL})$

m = 100.0 g

change in temperature, $\Delta T$ = 26.3 - 20.2 = 6.1 degree C

specific heat, s of water is 4.184 J per g per degree C.

$q=ms\Delta T$

where q is the heat energy. let's plug in the values in the equation:

$q=100.0g(4.184\frac{J}{g.^0C})6.1^0C$

q = 2552.24 J

energy of calorimeter = change in temperature*heat capacity of calorimeter

energy of calorimeter = $6.1^0C(\frac{12.1J}{^0C})$  

= 73.81 J

Total heat = 2552.24 J + 73.81 J = 2626.05 J

Let's convert it to kJ and for this we divide by 1000 since, 1000 J = 1 kJ

$2626.05J(\frac{1kJ}{1000J})$

= 2.626 kJ

0.05 moles of copper sulfate are used so the $\Delta H$ of the reaction = $\frac{2.626kJ}{0.05mol}$

= $\frac{52.52kJ}{mol}$

Since, there is an increase in temperature, the heat is released and so the sign of $\Delta H$ will be negative.

Hence, $\Delta H$ = -$\frac{52.52kJ}{mol}$

Answer 2

The $\rm \bold{\Delta H}$ of the process is 2.626 kJ/0.05 mol which is equal to 52.52 kJ/mol.

The heat capacity formula,

$\rm \bpld{ Q = mc \Delta T}$

Where,

Q - energy

m - mass = 100 mL

c - specific heat capacity =  4.184 $\rm \bold { J/g/^oC}$

$\rm \bpld{ \Delta T}$ - temperature difference = $\rm \bold { 6.1 ^oC}$

Put the value on equation we get

Q =  2552.24 J

Energy of calorimeter = 73.81 J

Total heat = 2552.24 J + 73.81 J = 2626.05 J =  2.626 kJ

Hence, the $\rm \bold{\Delta H}$ of the process is 2.626 kJ/0.05 mol which is equal to 52.52 kJ/mol.

To know more about specific heat capacity, refer to the link:

https://brainly.com/question/11194034?referrer=searchResults