At the start of a reaction, there are 0.0249 mol N2, 3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a 3.50 L reaction vessel at 37

Question

At the start of a reaction, there are 0.0249 mol N2,
3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a
3.50 L reaction vessel at 375°C. If the equilibrium constant, K, for the reaction:
N2(g) + 3H2(g)= 2NH3(g)
is 1.2 at this temperature, decide whether the system is at equilibrium or not. If it is not, predict in which direction, the net reaction will proceed.​

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Hải Đăng 5 months 2021-08-09T11:38:35+00:00 1 Answers 40 views 0

Answers ( )

    0
    2021-08-09T11:40:08+00:00

    Answer:

    Explanation:

    The reaction is given as:

    N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}

    The reaction quotient is:

    Q_C = \dfrac{[NH_3]^2}{[N_2][H_2]^3}

    From the given information:

    TO find each entity in the reaction quotient, we have:

    [NH_3] = \dfrac{6.42 \times 10^{-4}}{3.5}\\ \\ NH_3 = 1.834 \times 10^{-4}

    [N_2] = \dfrac{0.024 }{3.5}

    [N_2] = 0.006857

    [H_2] =\dfrac{3.21 \times 10^{-2}}{3.5}

    [H_2] = 9.17 \times 10^{-3}

    Q_c= \dfrac{(1.834 \times 10^{-4})^2}{(0.0711)\times (9.17\times 10^{-3})^3} \\ \\ Q_c = 0.6135

    However; given that:

    K_c = 1.2

    By relating Q_c \ \ and  \ \ K_c, we will realize that Q_c \ \ <  \ \ K_c

    The reaction is said that it is not at equilibrium and for it to be at equilibrium, then the reaction needs to proceed in the forward direction.

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