1.A 312 g sample of a metal is heated to 330.988 °C and plunged into 200 g of wa

Question

1.A 312 g sample of a metal is heated to 330.988 °C and plunged into 200 g of water at a temperature of 5.864 °C. The final temperature of the water is 75.041 °C. Assuming water has a specific heat capacity of 4.184 J/g °C, what is the specific heat capacity of the metal sample, in J/g °C)? Assume no heat loss to the surroundings.??

2.Consider the general reaction between a metal in its standard state and an acid (HA), to give the salt and hydrogen gas:

M(s) + 2 HA(aq) ? MA2(aq) + H2(g)

If the standard enthalpy of formation for HA is -101.761 kJ/mol and that of MA2(aq) is 365.280 kJ/mol, what is the standard enthalpy of reaction, in kJ?

3. coffee cup calorimeter is prepared, containing 100.000 g of water (specific heat capacity = 4.184 J/g K) at initial temperature 80.000 C. A salt weighing 2.918 g is quickly added. The salt has a molar mass of 320.889 g/mol. The final temperature of the solution is 57.849 C.

Assume no heat loss to the surroundings. Assume the specific heat capacity of the solution is equal to that of pure water, and that the mass of the solution is equal to the mass of the solid plus the mass of water in the calorimeter.

What is the molar heat of solution for the salt, in kJ/mol?

Explanation / Answer

A

Here the metal sample looses heat and the water gains heat. Under adiabatic conditions, heat lost by metal= mass of metal* specific heat of metal* change in temperature = heat gained by water= mass of water* specific heat of water* change in temperature

312* cp*(330.988-75.041)= 200*4.184*(75.041-5.864), Cp is the metal specific heat

Cp= 0.724 J/gm.deg.c

2.

M(s) + 2 HA(aq) ? MA2(aq) + H2(g), H2 is being gas, its standard enthalpy of formation =0 and that of metal=0

standard enthalpy of reaction = sum of enthalpy of formation of products- sum of enthalpy of formation of products

=1* standard enthalpy of formation of MA2+1* standard enthalpy of formation of H2- (1* standard enthalpy of formation of M+2*standard enthalpy of formation of HA)

=1*365.280-(2*(-101.761) =568.8 KJ

1,1,1 and 2 are coefficients of MA2, H2, M and HA respectively.

3.

coffee cup calorimeter is prepared, containing 100.000 g of water (specific heat capacity = 4.184 J/g K) at initial temperature 80.000 C. A salt weighing 2.918 g is quickly added. The salt has a molar mass of 320.889 g/mol. The final temperature of the solution is 57.849 C.

mass of the solution = mass of water+ mass of salt= 100+2.918= 102.918 gm

since there is a drop in temperature, the dissolution is endothermic and enthalpy change is +ve.

Enthalpy change= mass of the solution* specific heat* change in temperature = 102.918*4.184*(80-57.849) joules=9538 joules

mass of salt= 2.918 gm, molar mass =320.889 g/mole

moles of salt = mass/molar mass =2.918/320.889 =0.0090

enthalpy change per mole= 9538/0.0090 J/mole =1048883J/mole= 1048.883 Kj/mole

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