The plot of log 10 K v s 1 T gives a straight line. The intercept and slope respectively are (where K is equilibrium constant). [2026]

Question

The plot of $\log_{10} K vs \frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant). [2026]

Smart Achievers · Accepted Answer

(4)

$\log_{10} K = - \frac{Δ H °}{2.303 R T} + \frac{Δ S °}{2.303 R}$

$y-intercept = \frac{Δ S °}{2.303 R}$

$Slope = - \frac{Δ H °}{2.303 R}$

$Ans. (4) is correct.$

Solution

Q.
The plot of $\log_{10} K vs \frac{1}{T}$ gives a straight line. The intercept and slope respectively are (where K is equilibrium constant). [2026]

1 $\frac{2.303 R}{Δ H °}, \frac{2.303 R}{Δ S °}$

2 $- \frac{Δ H °}{2.303 R}, \frac{Δ S °}{2.303 R}$

3 $- \frac{Δ S ° R}{2.303}, \frac{Δ H ° R}{2.303}$

4 $\frac{Δ S °}{2.303 R}, - \frac{Δ H °}{2.303 R}$

Ans.
(4)

$\log_{10} K = - \frac{Δ H °}{2.303 R T} + \frac{Δ S °}{2.303 R}$

$y-intercept = \frac{Δ S °}{2.303 R}$

$Slope = - \frac{Δ H °}{2.303 R}$

$Ans. (4) is correct.$

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Solution

Q. The plot of log10K vs 1T gives a straight line. The intercept and slope respectively are (where K is equilibrium constant). [2026]

1 2.303RΔH°, 2.303RΔS°

2 -ΔH°2.303R, ΔS°2.303R

3 -ΔS°R2.303, ΔH°R2.303

4 ΔS°2.303R, -ΔH°2.303R

Ans. (4) log10K=-ΔH°2.303RT+ΔS°2.303R y-intercept=ΔS°2.303R Slope=-ΔH°2.303R Ans. (4) is correct.