What effect will decreasing the volume have the reaction? A) The reaction will s

Question

What effect will decreasing the volume have the reaction? A) The reaction will shift to the right B) The reaction will shift to the left C) The reaction will not change The reaction below takes place inside an instant cold pack when the ammonium nitrate is mixed with water. For this reaction, 2.00 g of NH_4NO_3 is dissolved in enough water to make a 30.0 mL solution. The initial temperature of the solution is 25.5 degree C and decreases to 21.2 degree C after the solid dissolves. NH_4NO_3 (s) rightarrow NH_4^+ (aq) + NO_3^- (aq) Assume that d_solution = 1.0 g/mL and C_s, solution = 4.184 J/(g middot degree C) Determine q_reaction A) +575.7 J B) -575.7 J C) -539.7 J D) +539.7 J Determine delta H_reaction. A) +23.03 kJ B) -23.03 kJ C) -21.59 kJ D) +21.59 kJ

Explanation / Answer

Answer for question 27.

Answer B: The reaction will shift to the left.

Explanation: In an equilibrium reaction when volume is decresed the pressure increases. According to Le Chatlier principle when a change occurrs in an equilibrium system the opposing events happen in order to bring back the equilibrium. Here when pressure is increased due to reduction in volume the reaction prroceeds in the direction where less number of moles of gas is generated.

Answer for question 28:

When a solute dissolved in a solvent the change in temperature of solution is known as enthalpy of solution or heat of solution or heat of reaction as given below and is calculated with the below formula.

qreaction = change in temperature x mass of solution x Csolution

In the above reaction change in temperature = - 4.3o C

mass of solution = 30 g ( d = 1 g/ml)

Csolution = 4.184 (J/g.o C)

Therefore qreactoin = 4.3 x 30 x 4.184

                            = 539.7

  Since the final temperature of reaction is less than initial temperature qreaction will be negative value i.e -539.7 J

So the answer is C = -539.7 J

H (kJ/mole) i. e enthalpy change  of a given reactant for the reaction is calculated as follows:

H = heat change(qreaction) /1000 ÷ moles

H = qreaction/1000 ÷ n

    = 539.7/1000 ÷ 0.025 (1 mole Ammonium nitrate = 80; 2 g equals to 0.025 mole)

    = 0.5397 ÷ 0.025

    = 21.59 kJ/mole

Since the above dissolution is exothermic reaction H will be negative and the answer will be C i.e -21.59 kJ/mole

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