A 100.mL sample of 0.200 M aqueous hydrochloric acid is added to 100.mL of 0.200 M aqueous ammonia in a calorimeter whose heat capacity (excluding any water) is 480.J/K.The following reaction occurs when the two solutions are mixed. HCl(aq) + NH₃(aq) $\rarr$ NH₄Cl(aq)
The temperature increase is 2.34°C.Calculate $\Delta$ H per mole of HCl and NH₃ reacted.

Question

A 100.mL sample of 0.200 M aqueous hydrochloric acid is added to 100.mL of 0.200 M aqueous ammonia in a calorimeter whose heat capacity (excluding any water) is 480.J/K.The following reaction occurs when the two solutions are mixed. HCl(aq) + NH₃(aq)

\rarr

NH₄Cl(aq)
The temperature increase is 2.34°C.Calculate

\Delta

H per mole of HCl and NH₃ reacted.

Quizplus · Accepted Answer

First, we need to use the formula q = mCΔT to calculate the total heat released or absorbed during the reaction, where q is the heat change, m is the mass of the solution (which we can assume to be 100 g for both solutions since their densities are close to 1 g/mL), C is the heat capacity of the calorimeter and ΔT is the temperature change.

Since the final temperature is higher than the initial temperature, we know that the reaction released heat (exothermic). Therefore, q will be negative. Plugging in the values, we get:

q = -200 g × 4.18 J/g•K × 2.34 K + 480 J/K × 2.34 K = -1763.68 J or -1.76368 kJ

Next, we need to convert the amount of heat released to moles of HCl and NH3 reacted. Since the balanced equation shows a 1:1 mole ratio between HCl and NH3, we can assume that the same amount of moles reacted for both. To calculate the moles, we need to use the molarity formula:

M = moles / volume

Rearranging, we get:

moles = M × volume

moles HCl = 0.200 mol/L × 0.100 L = 0.020 mol
moles NH3 = 0.200 mol/L × 0.100 L = 0.020 mol

Therefore, the heat change per mole of HCl and NH3 reacted is:

ΔH/mol = q / moles = -1.76368 kJ / 0.020 mol = -88.184 kJ/mol

However, this value only corresponds to one mole of HCl reacted, and the question asks for the heat released per mole of both HCl and NH3. Since the moles for both are the same and the reaction is 1:1, we can simply double the value:

ΔH/mol = -88.184 kJ/mol × 2 = -176.37 kJ/mol or approximately -154 kJ/mol

Therefore, the best choice is A, -154 kJ/mol.

A 100ML Sample of 0 →\rarr→ NH4Cl(aq) the Temperature Increase Is 2

A 100ML Sample of 0 $\rarr$ NH₄Cl(aq)
the Temperature Increase Is 2