Consider the following equilibrium:
2 H21g2 + S21g2 Δ 2 H2S1g2 Kc...

Question

Consider the following equilibrium:
2 H21g2 + S21g2 Δ 2 H2S1g2 Kc = 1.08 * 107at 700 C
(c) Calculate the value of Kc
if you rewrote the equation H21g2 + 12S21g2 Δ H2S1g2.

Accepted Answer

Hi everyone. So here we have the K. C. Value for the following reaction below. And this is 8.68 times 10 to the third at 27.9 degrees Celsius. Were asked what would be the K. C. Value if the following equation became the one below. We see that we have the same reactors in products, in both reactions of nitric oxide in auction, gas at the reactant and that nitrogen tetroxide as a product. But the coefficients in both reactions are different. In the first we actually have two moles of nitric oxide but only one In the second reaction, I have one more option gas in the first. Let me have one half moles of oxygen gas in the second And then we have one mole of dimension tetroxide in the 1st. 1/2 moles in the second. So this means that if we divide the first reaction by two, we get the second reaction. Sophie divide by two. This is the same as most climbed by one half. We raised the K. C. Value to the power of one half. Let me get Casey. It was 8.68 Times 10 to the 3rd. The power of one hat for K. C. We get money 3.2. Thanks for watching my video and I hope it was helpful

Consider the following equilibrium: 2 H21g2 + S21g2 Δ 2 H2S1g2 Kc = 1.08 * 107at 700 C (c) Calculate the value of Kc if you rewrote the equation H21g2 + 12S21g2 Δ H2S1g2.

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