Question

5. If the reaction in question 4 is A(aq) + B(aq) -AB(aq) and the molarity of A in solution A is 0.60 M and the molarity of B in solution B is M, calculate the enthalpy of reaction Afrr, for the formation of 1 mole of AB in solution 6. The addition of 20.0 J of heat to a 6.00 g sample of lead at 23.0 C caused the temperature to rise to 48.7 °C. What is the specific heat of lead? Define Hess's Law regaravess of the molripte stuges or stcps ofe a reactien the toto 7. entha The tau ts amantestatian tner emp enthalpy Calculate Δ// for the reaction 8. NO(g) + O(g) → NO2(g) given the following information: NO(g) + Os(g)- NO (8)+ 0(g) AH--198.9 k AH--142.3 kJ Δ1js 495.0 kJ 0)0) O2(g)-2 0(g)

Answer 1

6)

we have the formula q = m * s * dT

m = mass of the substance = 6 g

s = specific heat capcity of the substance

dT = change in temperature = (48.7-23) = 25.7

q = heat energy change = 20 J

20 = 6 * s * (25.7)

s = 0.1297 J/0C .g

7)

Hess law:

The heat energy change of a reaction is always constant. whether the reaction takes place in one step or in multiple steps.

8)

NO(g) +O3(g) ------> NO2(g) + O2(g) ; dH = -198.9 kJ -------1

O3(g) ------> 3/2O2(g) ; dH = -142.3 kJ ------2

O2(g) ------> 2 O(g) ; dH = 495 kJ -------3

reverse the equation 2 and 3

3/2O2(g) -----> O3(g) ; dH = 142.3 kJ -------4

2 O(g) ------> O2(g) ; dH = -495 kJ -------5

divide the equation 5 with factor 2

O(g) ------> 1/2O2(g) ; dH = -495/2 = -247.5 kJ-----6

Adding equations 1,4,6 then,

NO(g) +O3(g) ------> NO2(g) + O2(g) ; dH = -198.9 kJ -------1

3/2O2(g) -----> O3(g) ; dH = 142.3 kJ -------4

O(g) ------> 1/2O2(g) ; dH = -247.5 kJ-----6
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NO(g) + O(g) --------> NO2(g) ; dH = -304.1 kJ