Question

Calculate the in kJ/mole given that the following reaction has a

= -1219.26 kJ/mol

​2 ClF3(g) + 2 NH3(g) ----- > N2(g) + 6 HF(g) + Cl2(g)

​ Use data in appendix III

(b) The military uses FRH’s (Flameless ration heaters) to heat MRE’s (meals ready to eat). (Heat of formation of Mg( OH)2(Aqua) = -926.8 kJ/mole

One of the reactions used in an FRH is

​Mg(s) + 2 H2O(l) ------ > Mg(OH)2(aq) + H2(g)

​Using data in Appendix III of your textbook, calculate for this reaction

(c) Using results of part (b), determine how many grams of Mg(s) would be needed to heat 80.0 mL of water from 21.0oC to 80.0oC? Data: The specific heat of water is 4.184 J/(g deg C).

​(d) Consider two Styrofoam coffee cups. If cup #1 contains 80.0 mL of water at 96.0 oC and cup #2 contains 60.0 mL of water at 11.0 oC, what is the final temperature, in oC, when the contents of cup #1 are poured into cup #2? Ignore the heat capacity of the cup. Assume that the density of water is 1.00 g/mL.

Answer 1

Ans. 1

(a). For the reaction:

The energy of reaction is = -1219.26 KJ/mol;

Given that the energy involved in the formation of Mg(OH)₂ (aq) = -926 KJ/mol.

The remaining amount of energy will be released by flameless ration heaters (FRH).

(b) The energy released for the reaction used in FRH is:

Energy released for this reaction will be = -1219.26 - (-926) = -293.26 KJ /mole.

As 1 mole of Mg (s) is used in FRH reaction, so the energy for the reaction will be = -293.26 KJ/mole of Mg

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(c) First we will calculate the amount of energy required to heat 80 ml of water from 21 ^oC to 80 ^oC.

Energy required to heat 80 ml of water will be calculated by using following relation:

$Q = m C \Delta T$;

Here Q = Heat required

C = specific heat of water

$Q = 80 \times 4.18 \times (80-21) Joule$

Amount of heat required to change the temperature of water is : 19.729 KJ

Since, 1 mole of Mg is required to release the heat = 289.46 KJ

that is, 24 gram of Mg is required to release the heat = 289.46 KJ

Now we need to calculate the amount of Mg required to change the temperature of water:

So, gram of Mg required =

24 × 19.729 289.46 grams

= 1.63579 grams

$\approx 1.64 gms$, is the Mg required.

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(d) Here we have two coffee cups,

Given the content of cup #1 is poured into the content of cup #2, As water in cup #1 was at higher temperature and the water in the cup # 2 was at lower temperature. Upon mixing the temperature of cup #1 will decrease and the temperature of cup #2 increase.

To calculate the final temperature upon mixing, the principle to be applied is:  

The Heat released by cup # 1 = The Heat gained by Cup # 2

i.e. Change in Heat of water in cup #1 to the final temperature upon mixing = Change in Heat of water in cup # 2 to the final temperature upon mixing.

Hence :

$\Delta Q_1 = \Delta Q_2$

$\Delta Q_1 = m C \Delta T_1$;

$\Delta Q_2 = m C \Delta T_2$

Here Q = Heat required

C = specific heat of water

Assume final temperature upon mixing is T

Hence:

$80 \times C \times (96-T) = 60 \times C \times (T-11)$

$\Rightarrow 80 \times (96-T) = 60 \times (T-11)$

$\Rightarrow 384- 4T = 3T- 33$

$\Rightarrow 7T = 417$

$\Rightarrow T = 19.7 ^{o}C$

Hence the final temperature we calculated as: 19.7 ^oC. When the content of cup #1 is poured into the contents of cup # 2.

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