Part A A calorimeter contains 23.0 mL of water at 13.5 C . When 2.30 g of X (a s

Question

Part A

A calorimeter contains 23.0 mL of water at 13.5 C . When 2.30 g of X (a substance with a molar mass of 80.0 g/mol ) is added, it dissolves via the reaction

X(s)+H2O(l)X(aq)

and the temperature of the solution increases to 26.5 C .

Calculate the enthalpy change, H, for this reaction per mole of X.

Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(gC)], that density of water is 1.00 g/mL, and that no heat is lost to the calorimeter itself, nor to the surroundings.

Express the change in enthalpy in kilojoules per mole to three significant figures.

Part B

Consider the reaction

C12H22O11(s)+12O2(g)12CO2(g)+11H2O(l)

in which 10.0 g of sucrose, C12H22O11, was burned in a bomb calorimeter with a heat capacity of 7.50 kJ/C. The temperature increase inside the calorimeter was found to be 22.0 C. Calculate the change in internal energy, E, for this reaction per mole of sucrose.

Express the change in internal energy in kilojoules per mole to three significant figures.

****(A calorimeter is an insulated device in which a chemical reaction is contained. By measuring the temperature change, T, we can calculate the heat released or absorbed during the reaction using the following equation:

q=specific heat×mass×T

Or, if the calorimeter has a predetermined heat capacity, C, the equation becomes

q=C×T

At constant pressure, the enthalpy change for the reaction, H, is equal to the heat, qp; that is,

H=qp

but it is usually expressed per mole of reactant and with a sign opposite to that of q for the surroundings. The total internal energy change, E (sometimes referred to as U), is the sum of heat, q, and work done, w:

E=q+w

However, at constant volume (as with a bomb calorimeter) w=0 and so E=qv)

Explanation / Answer

a)

dHrxn = -Q/n

n = mass/MW = 2.3/80 = 0.02875

Q = m*cp*dt

Q = (2.3+23)*(4.184)*(26.5-13.5) = 1376.1176 J

Hrxn = -(1376.1176 )/0.02875 = 47864.96 J/mol

Hrxn = 47.9 kJ/mol

B)

m=10 g sucrose

mol = mass/MW = 10/342.2965 = 0.02921

Q = C*dT

Q = (7.5)(22) = 165 kJ

then

Hrxn = -Q/n = (-165)/0.02921 = -5648.75 kJ/mol

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