would anyone please answer this ? In a constant pressure calorimeter at standard
ID: 487454 • Letter: W
Question
would anyone please answer this ?
In a constant pressure calorimeter at standard conditions, 0.400 mol sodium metal is reacted with 432.0 mL of water as indicated in the following balanced thermochemical equation: Calculate BOTH Delta_r, H and Delta_r U for this reaction (at standard conditions), assuming ideal gas behavior and the following: the dry calorimeter apparatus has a total heat capacity of 678 J/degree C the calorimeter, water and sodium metal all have an initial temperature of 25.20 degree C and a final temperature of 33.20 degree C the aqueous solution that results has a total volume of 401.0 mL and a density of 1.04 g/mL the aqueous solution has the same specific heat as pure water: 4.184 J g^-1 K^-1 the calorimeter is perfectly insulated hints: Na is the limiting reactant! Possibly helpful equations: - q_rxn = q color = q_dry calor + q_soln = C_total Delta T + m_soln C_sp. soln Delta T Delta_r H = q_rxn/if at constant P Delta_r U = q_rxn/n_eqn mols LR if at constant V Delta_r H = Delta_r U + P Delta_r V = Delta_r U + Delta_r n_gas RT n_eqn mols I.R is the number of equation moles of the limiting reactant, where equation moles = actual moles/coefficient for that species in the balance chemical reactionExplanation / Answer
2 Na (s) + 2 H2O (l) -> 2 NaOH (aq) + H2 (g)
Initial moles of Na = 0.4
Initial volume of H2O = 432 mL
Initial mass of water = 432 * 1 g/mL = 432 g
Initial moles of water = 432/18 = 24
Hence, Na is the limiting reagent (LR)
Moles of NaOH formed = 0.4
Total mass of aqueous solution, M = Volume * Density = 401 mL * 1.04 g/mL
= 417.04 g
Cpsol = 4.184 J/g-K
Temperature change, dT = 33.2 – 25.2 °C = 8 °C
Qsol = M * Cpsol * dT
= 417.04 * 4.184 * 8 J
= 13959.16 J
Qdry cal = Heat capacity * dT
= 678 * 8 J
= 5424 J
-Qrxn = Qsol + Qdry cal
= (13959.16 + 5424) J = 19383.16 J
Qrxn = - 19383.16 J
Neqn moles of LR = 0.4 / 2 = 0.2
Hr = Qrxn / Neqn
= - 19383.16 / 0.2 = - 96915.8 J
T = 273 K
ng = Moles of H2 produced = 0.4 / 2 = 0.2
Hr = Ur + ng R T
- 96915.8 = Ur + 0.2 * 8.314 * 273
Ur = - 96915.8 - 453.94 J
= - 97369.8 J
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