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The decomposition of nitrosyl bromide is a second-order reaction: NOBr (g)   à    NO (g)   +    ½...

The decomposition of nitrosyl bromide is a second-order reaction: NOBr (g)   à    NO (g)   +    ½ Br2(g)

If the initial concentration of NOBr is 0.250 M, how much heat is evolved or absorbed per liter of solution in one minute? The rate constant is 0.80 L/(mol.s). The heats of formation are +82.2 kJ/mol for NOBr, +90.2 kJ/mol for NO, and +30.9 kJ/mol for Br2 as a gas.

Solutions

Expert Solution

Is a second order reaction so the equation you need to use is the following:

1/A = 1/Ao + kt

From here solve for A:

1/A = 1/0.250 + 0.80 * 60 = 52

A = 0.0192 mol/L

With the heat of formation, let's calculate rxn

rxn = prod - react

rxn = (90.2 + 30.9/2) - (82.2)

rxn = 23.45 kJ/mol

Finally, use the following expression to get the heat:

Q = rxn * [C]

Q = 23.45 kJ/mol * 0.0192 mol/L

Q = 0.45024 kJ/L or simply 450.24 J/L

Hope this helps

queen_honey_blossom answered 2 years ago
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