In a coffee-cup calorimeter, 1 mol naoh and 1 mol hbr initially at 22.5 oc (celsius) are mixed in 100g of water to yield the following reaction: naoh + hbr → na+(aq) + br-(aq) + h2o(l) after mixing the temperature rises to 83 oc. calculate the change in enthalpy of this reaction. specific heat of the solution = 4.184 j/(g oc) state your answer in kj with 3 significant figures. don't forget to enter the unit behind the numerical answer. the molecular weight of naoh is 40.0 g/mol, and the molecular weight of hbr is 80.9 g/mol.

The heat of reaction is -25.3 KJ/mol.

The equation of the reaction is; NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l)

The total mass of solution is 100g

The temperature rise =  83 oC - 22.5 oC = 60.5oC

Specific heat of the solution = 4.184 J/(g oC)

Number of moles = 1 mole

Given that the heat of reaction is calculated from;

ΔH = -(mcθ)/n

m =mass of solution

c = Specific heat of the solution

θ = temperature rise

n = number of moles of solution

ΔH = -(100g × 4.184 J/(g oC) ×  60.5oC)/1 mole

ΔH = -25.3 KJ/mol

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ΔH = -55.92 kJ

Explanation:

Step 1: Data given

1 mol NaOH and 1 mol HBr initially at 22.5 °C are mixed in 100g of water

After mixing the temperature rises to 83 °C

Specific heat of the solution = 4.184 J/g °C

Molar mass of NaOH = 40 G/mol

Molar mass of HBr = 80.9 g/mol

Step 2: The balanced equation

NaOH + HBr → Na+(aq) + Br-(aq) + H2O(l)

Step 3: mass of NaOH

Mass = moles * Molar mass

Mass NaOH = 1 * 40 g/mol

Mass NaOH = 40 grams

Step 4: Mass of HBr

Mass HBr = 1 mol * 80.9 g/mol

Mass HBr = 80.9 grams

Step 5: Calculate ΔH

ΔH = m*c*ΔT

ΔH= (100 + 40 + 80.9) * 4.184 * (83-22.5)

ΔH= 220.9 * 4.184 * 60.5

ΔH= 55916.86 J = 55.92 kJ

Since this is an exothermic reaction, the change in enthalpy is negative.

ΔH = -55.92 kJ

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.

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