Asked by sstt
The equilibrium constant for the reaction of hydrogen and nitrogen to form ammonia is Keq = 4.1 108 at 25 °C. A reaction was run and the equilibrium concentrations were found to be 1.0 10-3 M in both hydrogen and nitrogen and 0.020 M in ammonia. 0.010 M nitrogen was added.
a) Calculate Qeq for the system.
b) In which direction does the reaction shift?
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Write the reaction as N2 + 3H2 ⇌ 2NH3, so
Q = [NH3]^2 / ([N2][H2]^3).
After adding 0.010 M N2 the concentrations are [N2] = 0.011 M, [H2] = 1.0×10−3 M, [NH3] = 0.020 M.
a) Q = (0.020)^2 / (0.011·(1.0×10−3)^3)
= 4.0×10−4 / 1.1×10−11
≈ 3.6×10^7.
b) Keq = 4.1×10^8, so Q < Keq. The reaction will shift to the right (forward) to produce more NH3 until equilibrium is re-established.
Q = [NH3]^2 / ([N2][H2]^3).
After adding 0.010 M N2 the concentrations are [N2] = 0.011 M, [H2] = 1.0×10−3 M, [NH3] = 0.020 M.
a) Q = (0.020)^2 / (0.011·(1.0×10−3)^3)
= 4.0×10−4 / 1.1×10−11
≈ 3.6×10^7.
b) Keq = 4.1×10^8, so Q < Keq. The reaction will shift to the right (forward) to produce more NH3 until equilibrium is re-established.
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