NaOH(aq) + HNO_3(aq) rightarrow NaNO_3(aq) + H_2O(l) Delta H = -56.0 kJ 50.0 mL

Question

NaOH(aq) + HNO_3(aq) rightarrow NaNO_3(aq) + H_2O(l) Delta H = -56.0 kJ 50.0 mL of 2.00 M NaOH is mixed with 50.0 mL of 1.00 M HN03 in a Styrofoam cup calorimeter. The initial temperature of each solution is 23.0 degree C. What is the final temperature after reaction is complete? Which of the following reactions has a larger negative value of Delta H? Clearly explain your answer. Calculate energy transferred as work in the following reaction at 1 atm and 25 degree. N_2(g) + 3H_2(g) rightarrow 2NH_3(g)

Explanation / Answer

4)
mol of NaOH added = M*V = 2.00 M * 0.050 L = 0.10 mol
mol of HNO3 added = M*V = 1.00 M * 0.050 L = 0.05 mol

0.05 mol of each will react

heat released = number of mol * H
= 0.05 mol * 56.0 KJ/mol
=2.8 KJ
= 2800 J

volume of solution = 100 mL
assuming density is 1g/mL, mass = 100 g

Now use:
Q = m*C*(Tf-Ti)
2800 = 100*4.184*(Tf-23)
Tf = 29.7 oC

Answer: 29.7 oC

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