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Given the balanced chemical equation: Ba(NO3)2(aq) + Na2SO4(aq) → BaSO4(s) + 2 NaNO3(aq) A 0.30 L of 0.20 M Ba(NO3)2(aq) is mixed with excess Na2SO4(aq) to make 200. g of solution in a coffee-cup calorimeter and allowed to react completely. The temperature of the solution rises from 25.2 ºC to 27.1 ºC. Answer the question in each step to find ∆Hrxn for this reaction. Cs,soln = 4.18 J/g ºC

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Answer:

∆Hrxn = -26467 J/mol

Explanation:

First, you need to know the reaction is exothermic because the solution temperature increases.

The heat of reaction, Q, is defined as:

Q = -C×m×ΔT

Where C is 4,18J/g°C; m is mass (200g); and ΔT is (27,1°C - 25,2°C = 1,9°C)

Q = 4,18J/g°C × 200g × 1,9°C

Q = -1588J

Now, moles of reaction are 0,30L × (0,20mol / L) = 0,06mol

That means ∆Hrxn is:

∆Hrxn: -1588J / 0,06mol = -26467 J/mol

I hope it helps!

Answer:

∆Hrxn for this reaction= -26467 J/mol

Explanation:

  • Given informations:  

The reaction is exothermic because the solution temperature increases.

C =4,18J/g°C

mass =200g

ΔT =27,1°C - 25,2°C = 1,9°C

  • Formula:

Q = -C×m×ΔT

Q = 4,18J/g°C × 200g × 1,9°C

Q = -1588J

Moles of reaction =0,30L × (0,20mol / L) = 0,06mol

Therefore ,

∆Hrxn: -1588J / 0,06mol = -26467 J/mol

Thus, the answer is -26467.

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