Question

Question 1 2 pts Given the balanced chemical equation: Thermometer Ba(NO3)2(aq) + Na2SO4(aq) →BaSO4(s) + 2 NaNO3(aq) - Stirrer A 0.30 L of 0.15 M Ba(NO3)2(aq) is mixed with excess Na2SO4(aq) to make 200.0 g of solution in a coffee-cup calorimeter and allowed to react completely. The temperature of the solution rises from 23.5 °C to 24.9 °C. Answer the question in each step to find Hrxn for this reaction. Cs,soln = 4.18 J/g °C Cover Step 1: The reaction is exothermic because the solution temperature increases. The heat absorbed by the solution Reaction mixture (asoln) is (Select] Styrofoam cups Step 2: The heat associated with the reaction (Grxn) is [Select] Step 3: Use 9rxn and the stoichiometric relations in the balanced chemical equation to calculate AHrxn. AHrxnis [Select] Next

Answer 1

Question 1

Number of moles of
Ba(NO3)2
in 0.30 L of 0.15 M
Ba(NO3)2
solution

0.15 mol/L x 0.30 L = 0.045 mol

Step 1

The reaction between barium nitrate and sodium sulfate is exothermic due to increase in the temperature of solution.

Calculate the heat absorbed by solution

Isoln = Mass of solution x Cs, soln X AT

$q_{soln}= \text { 200.0 g } \times \text { 4.18 }\frac{J}{g^oC} \times \left ( 24.9 \ ^oC-23.5 \ ^oC \right )$

$q_{soln}= \text { 200.0 g } \times \text { 4.18 }\frac{J}{g^oC} \times 1.4 \ ^oC$

Step 2

The heat associated with the reaction is

$q_{rxn}=-q_{soln}$

Step 3:

$\Delta H_{rxn} =\frac{ q_{rxn}}{ \text { number of moles of barium nitrate}}$

$\Delta H_{rxn} =\frac{-1170.4 \ J}{0.045 \ mol}$

$\Delta H_{rxn} =-26009 \ \frac{ J}{ mol}$

Convert unit from J/mol to kJ/mol

$\Delta H_{rxn} =-26009 \ \frac{ J}{ mol} \times \frac{1 \ kJ}{1000 \ J}$

$\Delta H_{rxn} =-26 \ \frac{ kJ}{ mol}$

Hence, the enthalpy change for the reaction is $-26 \ \frac{ kJ}{ mol}$ .