I need help with 2 and 5 please Predict some common process to be exothermic or

Question

I need help with 2 and 5 please Predict some common process to be exothermic or endothermic xml determine their sign of Delta H. Apply Delta H_Des in thermochemical stoichiometric calculations. Calculate Delta H_Des based on calorimetry data. Predict if the following processes are exo- or endothermic. Rationalize sign for Delta H. Condensation of water steam. Combustion of wood. Nuclear fusion in the Sun. Two solutions were both at 20.0 degree C initially. When they are mixed, the temperature of the solution became 18.0 degree C. Is the reaction between the two solutions endo- or exothermic? What is the sign of its DeltaH_Bor? Briefly support your conclusion. An unknown metal is cither aluminum, iron or lead. If 150. g of this metal at 150.0 degree C. was placed in a calorimeter that contains 200. g of water at 25.0 degree C and the final temperature was found to be 34.3 degree C after thermal equilibrium was achieved. Assume heat was only transferred between water and metal, what is the identify of this metal? Some specific heat values of metals and water given below may be useful. Specific heats, J/(g degree C): Fe (0.449) Pb (0.128) A1 (0.903) H_2O (4.184) H_2A and BeOH are acid and base and they react according to the following balanced equation: H_2A(aq) + 2 BOH(aq) rightarrow B_2A(aq) + 2 H_2O(l) If 0.20 moles of BOH reacted with excess amount of H_2A solution and 1500. J of heat was released by the reaction, what is the Delta H_nes for the reaction as written above? NH_4NO_3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the temperature from 25.0 degree C to 5.0 degree C, how many grams of NH_4NO_3 should we use for every 100.0 g of water in the cold pack? Assume no heat was lost outside of cold pack, and the specific heat of the resulted solution was the same as water, or 4.184 J/(g^* degree C).

Explanation / Answer

Q. 2

Given :

Initial T = 20.0 deg C , Final T = 18.0 deg C

Solution:

Calculation of delta T:

Delta T = Tf – Ti = 18.0 deg C – 20.0 deg C = -2 deg C

This shows that temperature of the system is lowered. The amount of lowered T is taken by the reaction to proceed.

If the reaction absorbs the heat then the reaction is endothermic.

So given reaction is endothermic.

And the sign convension for endothermic reaction is always +ve

Q. 5

Given :

Heat of solution of NH4NO3 = 25.7 kJ/mol

Delta T = Final T - Initial T = 5 deg C – 20.0 deg c = - 15.0 deg C

Mass of water = 100.0 g

We have to find q of the system

q = m x c x delta T

m is mass of water in g , C is specific heat , delta T is change in T

q = 100.0 g x 4.184 J / (deg C g ) x (-15 deg C )

= -6276 J

q = - 6276 J

we know delta Hrxn = - q/n

n = - ( - 6276 J ) / ( 25700 J/mol)

= 0.244 mol

Mass of NH4NO3 = 0.244 mol x molar mass of NH4

= 0.244 mol x 80.0434 g/mol

= 19.55 g

Mass of NH4NO3 required = 19.55 g

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