Use the table of thermodynamic data in your text, a Chemistry/Physics CRC, or th

Question

Use the table of thermodynamic data in your text, a Chemistry/Physics CRC, or the internet (but be careful with data off the internet), to calculate the molar enthalpy of the following reactions:

1) An aqueous solution of sodium hydroxide reacts with an equeous solution of hydrochloric acid, yielding water.

2) An aqueous solution of sodium hydroxide reacts with an aqueous solution of ammonium chloride, yielding aqueous ammonia, NH3, and water.

3) An aqueous solution of hydrochloric acid reacts with aqueous ammonia, NH3, yielding aqueous ammonium chloride.

Explanation / Answer

1) An aqueous solution of sodium hydroxide reacts with an equeous solution of hydrochloric acid, yielding water.

Heat of reaction=1*(-407.27)+ 1* (-285.83) - 1*( -167) -1 *(-469.15)= -407.27 -285.83 +167 +469.15= -56.95kJ/mol

Hrxn =  -56.95kJ/mol

2) An aqueous solution of sodium hydroxide reacts with an aqueous solution of ammonium chloride, yielding aqueous ammonia, NH3, and water.

NaOH + NH4Cl ---> NH3 + H2O + NaCl

Hrxn = Hproducts - Hreactants

[1Hf(NH3 (g)) + 1Hf(H2O (l)) + 1Hf(NaCl (aq))] - [1Hf(NH4Cl (aq)) + 1Hf(NaOH (aq))]

[1(-46.11) + 1(-285.83) + 1(-407.25)] - [1(-299.66) + 1(-470.09)]

30.56 kJ     (endothermic)

3) An aqueous solution of hydrochloric acid reacts with aqueous ammonia, NH3, yielding aqueous ammonium chloride.

HCl + NH3 --> NH4Cl

H = [1moleNH4Cl( - 314.55 kJ/mole)] – [1 mole NH3( -45.90 kJ/mole) + 1 mole HCl( -92.31 kJ/mole)]

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