Asked by qwerty
N2 + 3H2 ⇌ 2NH3, equilibrium constant = 6.0 x 10^-2
find the reaction quotient and the direction in which the system will shift to reach equilibrium when [NH3] initial = 2.30 x 10^-4 M, [N2] initial = 3.54 x 10^-2 M, [H2] initial = 1.50 x 10^-2 M
find the reaction quotient and the direction in which the system will shift to reach equilibrium when [NH3] initial = 2.30 x 10^-4 M, [N2] initial = 3.54 x 10^-2 M, [H2] initial = 1.50 x 10^-2 M
Answers
Answered by
DrBob222
Q = (NH3)^2/(N2)(H2)^3
Plug in the numbers.
Q = (2.3E-4)^2/(3.54E-2)(1.5E-2)^3 = My calculator is on the blink but I'm guessing about 0.4. You should confirm that.
So K = 0.06. Thus Q is too large (if that 0.4 is a good number) which means numerator (product side) is too large and/or denominator (reactant side) is too small. How can the reaction shift to make product smaller and reactants larger? left to make Q smaller; right to make Q larger. Hint: it won't shift to the right.
Plug in the numbers.
Q = (2.3E-4)^2/(3.54E-2)(1.5E-2)^3 = My calculator is on the blink but I'm guessing about 0.4. You should confirm that.
So K = 0.06. Thus Q is too large (if that 0.4 is a good number) which means numerator (product side) is too large and/or denominator (reactant side) is too small. How can the reaction shift to make product smaller and reactants larger? left to make Q smaller; right to make Q larger. Hint: it won't shift to the right.
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