You dissolve 0.1 mol of ammonium nitrate in 100mL of H2O. The temperature of the

Question

You dissolve 0.1 mol of ammonium nitrate in 100mL of H2O. The temperature of the solution changes from 22C to 15C. The specific heat capacity for water is 4.184J / gK.

A. Is this reaction endothermic or exothermic?

B. Calculate the heat, q, released or absorbed in the dissolution of the reaction.

C. Calculate the enthalpy change, dH(rxn) (kJ / mol), for dissolving 1 mol of ammonium nitrate.

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I chose this to be endothermic based on:

q = ms(dT).

dT = 15C - 22C = -7C or 266K

NH4NO3 has a molar mass of 80.05g and 0.1 mol has about 8g so:

q = (8g)(4.184J/gK)(266K) = 8903J or 8.9 x 10^3 J

How do I calculate the Enthalpy change (dHrxn) for 1 mol?

Explanation / Answer

A. Your answer is correct. ENDOTHERMIC.

B. q=mass*specific heat*change in temperature=100*4.184*7=2.9288 kJ energy absorbed

C. The above value is for 0.1 mol.

For 1mol, multiply by 10 to get -29.288kJ/mol

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