Question

A student mixed 25.0cm³ of 0.350 mol dm⁻³ sodium hydroxide solution with 25.0 cm³ of 0.350 mol dm⁻³ hydrochloric acid. The temperature rose by 2.50°C. Assume that no heat was lost to the surroundings.

The final mixture had a specific heat capacity of 4.20 J cm 3 K⁻¹.
What is the molar enthalpy change for the reaction?
(a) -150 kJ mol⁻¹
(b)-60.0 kJ mol⁻¹
(c) -30.0 kJ mol⁻¹
(d)-0.150 kJ mol⁻¹

Answer 1

The molar enthalpy change for the reaction is -0.150 kJ mol⁻¹.

Step-by-step explanation:

The molar enthalpy change for the reaction can be calculated using the equation q = mcΔT, where q is the heat produced or absorbed, m is the mass of the solution, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, the solution is a mixture of 25.0 cm³ of 0.350 mol dm⁻³ sodium hydroxide and 25.0 cm³ of 0.350 mol dm⁻³ hydrochloric acid. The temperature change is 2.50°C and the specific heat capacity is 4.20 J cm³ K⁻¹.

Using these values, we can calculate the molar enthalpy change:

q = mcΔT

q = (50.0 cm³)(4.20 J cm³ K⁻¹)(2.50°C)

q = 525 J

Since the reaction involves 0.350 mol of both sodium hydroxide and hydrochloric acid, the molar enthalpy change is

-525 J / (0.350 mol + 0.350 mol) = -750 J mol⁻¹ = -0.750 kJ mol⁻¹

Therefore, the correct answer is (d) -0.150 kJ mol⁻¹.