When 0.800 grams of NaOH is dissolved in 100.0 grams of water, the temperature o

Question

When 0.800 grams of NaOH is dissolved in 100.0 grams of water, the temperature of the solution increases from 25.00 to 27.06 degree C. Calculate the enthalpy of dissolution of NaOH in kJ/mol. (The specific heat of solution is 4.18 J/g degree C.) Enter your answer with no units. Make sure to indicate the sign (negative or positive). What is Delta H degree_rxn for the following reaction? Enter your answer with no units. Make sure to indicate the sign (positive or negative). C_6H_6(I) + 15/2 O_2(g) rightarrow 6CO_2(g) + 3H_2O(I) 6C(graphite) + 3H_2(g) rightarrow C_6H_6(..) Delta H degree_rxn = +49.0 kj/mol C(graphite) + O_2(g) rightarrow CO_2(g) Delta H degree_rxn = -393.5 kj/mol H_2(g) + 1/2 O_2(g) rightarrow H_O() Delta H degree_rxn = -285.8 kj/mol

Explanation / Answer

First calculate the total amount of heat as follows:

Q = mcdT

Here m total mass of solution , 0.800 +100.0 = 100.8 g

C = specific heat of solution, 4.18 J/ g-C

dT = temperature change , T2-T1

27.06-25.00= 2.06 C

Therefore;

Q = mcdT

Q = 100.8 * 4.18 J/ g-C *2.06 C

= 867.97 J

= 0.88 kJ

Now calculate the number of moles of NaOH

Amount of NaOH / Molar mass

= 0.88 g /39.997 g/ mole

= 0.02 moles

Now calculate the enthalpy of dissociation of NaOH per mole divide the total amount of heat by number of moles:

0.88 kJ /0.02 mole

=+ 44 KJ/ mole

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