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A 1.00 x 102mL sample of 0.200 M aqueous hydrochloric acid is added to 1.00 x 102mL of 0.200 M aqueous ammonia in a constant-pressure calorimeter of negligible heat capacity. The following reaction occurs when the two solutionsare mixedHCl(aq)+ NH3(aq)--->NH4Cl(aq)The temperature increase is 2.34°C. Calculate heat change of the reaction per mole of HCl reacted.Assume that the densities and specific heats of the solutions are the same as for water (1.00 g/mL and 4.184 J/g · °C, respectively)

User Audry
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1 Answer

5 votes

Answer:

ΔH = -98 kJ/mol

Step-by-step explanation:

To calculate the heat change of the reaction:

HCl(aq) + NH₃(aq) → NH₄Cl(aq)

0.2M 0.2M ΔT=2.34°C

1x10²mL 1x10²mL

We need to use the next equation:


q = mc \Delta T
(1)

where q: the amount of heat energy lost or gained, m: the mass of the substance, c: the specific heat capacity of the substance and ΔT: the change in temperature of the substance

Assuming that the densities of the solutions are the same as for water, we can determine the mass of the solution:


d = \frac {m}{V}

where d: density, m: mass and V: volume of solution = 100 + 100 = 200mL


m = d \cdot V = 1 \frac {g}{mL} \cdot 200mL = 200g

Now, using the calculated mass in equation (1), and assuming that the specific heats of the solutions are the same as for water, we can find heat change of the reaction:


q = 200g \cdot 4.184 \frac {J}{g \cdot ^(\circ)C} \cdot 2.34^(\circ)C


q = 1.96 \cdot 10^(3)J

This heat is negative because is the heat lost by the reacting HCl and NH₃ and gained by the water, so:


q = - 1.96 \cdot 10^(3)J

To calculate the heat change of the reaction per mole of HCl, we need to divide the heat change by the number of moles, which is called the enthalpy of reaction:


\Delta H = \frac {q}{moles}


\Delta H = \frac {q}{moles}


\Delta H = \frac {-1.96 \cdot 10^(3) J}{0.2 (mol)/(L) \cdot 0.1L}


\Delta H = -98 \cdot 10^(3) (J)/(mol) = -98 \frac {kJ}{mol}

So, the heat change of the reaction per mole of HCl reacted, often called enthalpy of reaction, is ΔH = -98 kJ/mol.

Have a nice day!

User NolanDC
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