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    couls someone explain how the gradient of the graph of delta g against t is equal to delta s ?? i dont get why delta h disappears from the equation ?
    thanks !!
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    (Original post by HG1)
    couls someone explain how the gradient of the graph of delta g against t is equal to delta s ?? i dont get why delta h disappears from the equation ?
    thanks !!


    There is the equation. A linear graph is usually of the form y = mx + c (where the c represents the y intercept (where x = 0), m is the gradient, x is the thing on the x axis and y the y axis).

    The thing above looks slightly different - it is y = c + mx where m = - delta s. The c (delta h) is not part of the gradient term so it disappears - the gradient of a straight line is just the thing that is next to the 'x'.
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    (Original post by SeanFM)


    There is the equation. A linear graph is usually of the form y = mx + c (where the c represents the y intercept (where x = 0), m is the gradient, x is the thing on the x axis and y the y axis).

    The thing above looks slightly different - it is y = c + mx where m = - delta s. The c (delta h) is not part of the gradient term so it disappears - the gradient of a straight line is just the thing that is next to the 'x'.
    Thanks for the reply its helping , could you show the gibbs equation is written out in terms of y=mc+c ??
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    (Original post by HG1)
    Thanks for the reply its helping , could you show the gibbs equation is written out in terms of y=mc+c ??
    I did so in my post just to repeat - if a line is given by the equation y = mx + c, then m is the gradient.

    The equation is of the form y = mx + c, where y = delta G, c = delta H, x = T and m = delta S. If you want it to look exactly like y = mx+c then you just rearrange it so that delta G = -delta S * T + delta H.
 
 
 
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