Nitric acid can be formed in two steps from the atmospheric gases nitrogen and oxygen, plus hydrogen prepared by reforming natural gas. In the first step, nitrogen and hydrogen react to form ammonia: N2(g) + 3 H,(9) - 2 NH,(g) AH=-92. kJ In the second step, ammonia and oxygen react to form nitric acid and water: NH,(g) + 20,(g) - HNO,(g) + H,O(g) AH=-330. kJ Calculate the net change in enthalpy for the formation of one mole of nitric acid from nitrogen, hydrogen and oxygen from these reactions. Round your answer to the nearest kJ.

q represents the change in heat of a reaction or substance. The units of q are Joules (J) or kilojoules (kJ). For a well understood substance, we can calculate the specific heat which relates the mass of the substance in grams to the amount of heat in J required to raise that amount a certain temperature.  We call that specific heat.  The specific heat (Cs) of water is stated in the following way:  C s ( H 2 O ) = 4.186 J g ∘ C This means that it takes 4.186 J of energy to raise the temperature of 1g of water by 1ºC.  The relationship between q and Cs for a given substance is  where m is mass (g) and ∆T is the change in temperature in ºC (final temp - initial temp).  Calculate the heat change (q) in J when 4.297 grams of water is cooled from 3.967 ºC to -2.902 ºC. Enter your answer with one decimal place (tenths).  Pay careful attention to the sign of your answer.  The sign of ∆T is always the same as the sign for q.

When 6.54 grams of Zn is placed in 500.0 mL of 1.00 M CuSO4(aq) in a coffee cup calorimeter, it reacts completely to displace copper. The temperature of the solution rises from 20.0˚C to 30.4˚C. Assume the coffee cup itself gains no heat and that the solution has the same density (1.00 g/mL) and specific heat (4.184 J/g˚C) as pure water. (a) How much heat does the solution gain during this reaction? (in J)

How many moles of butane, C4H10(l), must be burned to produce 234 kJ of heat under standard state conditions? (The heat of combustion for butane is −2855.7 kJ/mol.)

4. When 1.00 L of 1.05 M Ba(NO3)2 solution is mixed with 1.00 L of 1.10 M NazSO4 solution at 25.0°C in a coffee-cup calorimeter. The reaction is Ba(NOs)2(aq) + NazSO4(aq) → BaSO4(s) + 2NANO:(aq). The final temperature of the mixture increases to 28.1°C. Calculate the enthalpy change per mole for this process. (Assuming C=4.18 J °C' g' and density of the final solution is 1.05 g/ml).

MN.19.

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