The reaction of nitrogen monoxide(g) with hydrogen(g) to form nitrogen(g) and water(l) proceeds as follows: 2NO(g)...
The reaction of nitrogen monoxide(g) with hydrogen(g) to form nitrogen(g) and water(l) proceeds as follows: 2NO(g) + 2H2(g) N2(g) + 2H2O(l) When 6.39 g NO(g) reacts with sufficient H2(g), 80.1 kJ is evolved. Calculate the value of deltaH for the chemical equation given.
Homework Answers
Molar mass of NO,
MM = 1*MM(N) + 1*MM(O)
= 1*14.01 + 1*16.0
= 30.01 g/mol
mass(NO)= 6.39 g
use:
number of mol of NO,
n = mass of NO/molar mass of NO
=(6.39 g)/(30.01 g/mol)
= 0.2129 mol
Q = -80.1 KJ
This is when 0.2129 mol of NO reacts.
In reaction, 2 mol of NO is reacting.
So,
Delta H = 2 * Q / number of mol
= 2 * (-80.1 KJ) / 0.2129 mol
= -752 KJ/mol
Answer: -752 KJ/mol
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