A) When 1 mole of CH 4 (g) reacts with O 2 (g) to form CO 2 (g) and H 2 O(g) acc

Question

A) When 1 mole of CH4(g) reacts with O2(g) to form CO2(g) and H2O(g) according to the following equation, 802 kJ of energy areevolved.

Is this reaction endothermic or exothermic? _________endothermicexothermic

What is the value of q?______________ kJ

B)

When a solid dissolves in water, heat may be evolved or absorbed. The heat of dissolution(dissolving) can be determined using a coffee cup calorimeter.

In the laboratory a general chemistry student finds that when 6.18 g of NH4Br(s) are dissolved in 105.40 g of water, the temperature of the solution drops from 23.21 to 20.56°C.

The heat capacity of the calorimeter (sometimes referred to as the calorimeter constant) was determined in a separate experiment to be 1.80 J/°C.

Based on the student's observation, calculate the enthalpy of dissolution of NH4Br(s) in kJ/mol.

Assume the specific heat of the solution is equal to the specific heat of water.

Hdissolution = ____________________kJ/mol

Explanation / Answer

Answer – A) We are given reaction with change in enthalpy –

CH4(g) + 2O2(g) ----> CO2(g) + 2H2O(g) Ho = 802 kJ

We know when reaction evolve the energy in the surrounding then it is exothermic reaction.

So this reaction exothermic

We know,

Ho = -q

       q = - (802 kJ)

            = - 802 kJ.

B) We are given, mass of NH4Br = 6.18 g , mass of water = 105.40 g, ti = 23.21o C, tf = 20.56oC , heat capacity of the calorimeter = 1.80 J/oC

We know the density of water is 1.0 g/mL

Now we need to calculate the loss of heat by water

We know, heat loss by water = heat gain by NH4Br + heat gain by calorimeter

Heat loss by water

Heat, q = m* C * t

             = 105.40 g * 4.184 J/goC * (20.56-23.21)oC

             = -1168.6 J

Heat gain by calorimeter

Heat, q = C *t

           = 1.80 J/oC * (20.56-23.21)oC

           = - 4.77 J

So, heat of dissolution of NH4Br = heat loss by water – heat gain by colorimeter

                                                      = -1168.6 J – (4.77 J)

                                                       = -1163.8 J

                                                       = -1.164 kJ

We know, H = -q

So, H = 1.164 kJ

Now we need to calculate moles of 6.18 g of NH4Br

Moles of NH4Br = 6.18 g / 97.94 g.mol-1

                             = 0.0631 moles

So, H dissolution of NH4Br = 1.164 kJ / 0.0631 moles

                                                 = 18.44 kJ/mol

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